NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections are provided here to enhance the Maths exam preparation for the students as well as to score better marks. The expert Maths teacher have developed these NCERT Solutions for Chapter 11 in accordance with the updated syllabus of CBSE Class 11. Moreover, the solutions have been developed in various ways to match the understanding level of the students, so that they can grasp the difficult Maths concepts with ease.

Chapter 11 aggregates some of the quintessential topics such as Introduction to Conic Sections, Sections of a Circle as well as Circle, Parabola, Hyperbola and Ellipse. So, in order to develop a thorough insight into all these topics, students can rely on NCERT Solutions. Using these solutions, the students can practise a more significant number of challenging questions from the chapter to enhance their question-solving skills.

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EXERCISE 11.1 PAGE NO: 241

**In each of the following Exercise 1 to 5, find the equation of the circle with
1. Centre (0, 2) and radius 2**

**Solution:**

Given:

Centre (0, 2) and radius 2

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x â€“ h)^{2 }+ (y â€“ k)^{2 }= r^{2}

So, centre (h, k) = (0, 2) and radius (r) = 2

The equation of the circle is

(x â€“ 0)^{2 }+ (y â€“ 2)^{2 }= 2^{2}

x^{2 }+ y^{2 }+ 4 â€“ 4y = 4

x^{2 }+ y^{2}Â â€“ 4y = 0

âˆ´ The equation of the circle is x^{2 }+ y^{2}Â â€“ 4y = 0

**2. Centre (â€“2, 3) and radius 4**

**Solution:**

Given:

Centre (-2, 3) and radius 4

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x â€“ h)^{2 }+ (y â€“ k)^{2 }= r^{2}

So, centre (h, k) = (-2, 3) and radius (r) = 4

The equation of the circle is

(x + 2)^{2}Â + (y â€“ 3)^{2}Â = (4)^{2}

x^{2}Â + 4x + 4 + y^{2}Â â€“ 6y + 9 = 16

x^{2}Â + y^{2}Â + 4x â€“ 6y â€“ 3 = 0

âˆ´ The equation of the circle is x^{2}Â + y^{2}Â + 4x â€“ 6y â€“ 3 = 0

**3. Centre (1/2, 1/4) and radius (1/12)**

**Solution:**

Given:

Centre (1/2, 1/4) and radius 1/12

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x â€“ h)^{2 }+ (y â€“ k)^{2 }= r^{2}

So, centre (h, k) = (1/2, 1/4) and radius (r) = 1/12

The equation of the circle is

(x â€“ 1/2)^{2} + (y â€“ 1/4)^{2} = (1/12)^{2}

x^{2} â€“ x + Â¼ + y^{2} â€“ y/2 + 1/16 = 1/144

x^{2} â€“ x + Â¼ + y^{2} â€“ y/2 + 1/16 = 1/144

144x^{2}Â â€“ 144x + 36 + 144y^{2}Â â€“ 72y + 9 â€“ 1 = 0

144x^{2}Â â€“ 144x + 144y^{2}Â â€“ 72y + 44 = 0

36x^{2}Â + 36x + 36y^{2} â€“ 18y + 11 = 0

36x^{2}Â + 36y^{2}Â â€“ 36x â€“ 18y + 11= 0

âˆ´ The equation of the circle is 36x^{2}Â + 36y^{2}Â â€“ 36x â€“ 18y + 11= 0

**4. Centre (1, 1) and radius âˆš2**

**Solution:**

Given:

Centre (1, 1) and radius âˆš2

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x â€“ h)^{2 }+ (y â€“ k)^{2 }= r^{2}

So, centre (h, k) = (1, 1) and radius (r) = âˆš2

The equation of the circle is

(x-1)^{2}Â + (y-1)^{2 }= (âˆš2)^{2}

x^{2}Â â€“ 2x + 1 + y^{2}Â -2y + 1 = 2

x^{2}Â + y^{2}Â â€“ 2x -2y = 0

âˆ´ The equation of the circle is x^{2}Â + y^{2}Â â€“ 2x -2y = 0

**5. Centre (â€“a, â€“b) and radius âˆš(a ^{2} â€“ b^{2})**

**Solution:**

Given:

Centre (-a, -b) and radius âˆš(a^{2} â€“ b^{2})

Let us consider the equation of a circle with centre (h, k) and

Radius r is given as (x â€“ h)^{2 }+ (y â€“ k)^{2 }= r^{2}

So, centre (h, k) = (-a, -b) and radius (r) = âˆš(a^{2} â€“ b^{2})

The equation of the circle is

(x + a)^{2}Â + (y + b)^{2}Â = (âˆš(a^{2} â€“ b^{2})^{2})

x^{2}Â + 2ax + a^{2}Â + y^{2}Â + 2by + b^{2}Â = a^{2}Â â€“ b^{2}

x^{2}Â + y^{2}Â +2ax + 2by + 2b^{2}Â = 0

âˆ´ The equation of the circle is x^{2}Â + y^{2}Â +2ax + 2by + 2b^{2}Â = 0

**In each of the following Exercise 6 to 9, find the centre and radius of the circles.**

6. (x + 5)^{2}Â + (y â€“ 3)^{2}Â = 36

**Solution:**

Given:

The equation of the given circle is (x + 5)^{2}Â + (y â€“ 3)^{2}Â = 36

(x â€“ (-5))^{2} + (y â€“ 3)^{2}Â = 6^{2} [which is of the form (x â€“ h)^{2}Â + (y â€“ k )^{2}Â = r^{2}]

Where, h = -5, k = 3 and r = 6

âˆ´ The centre of the given circle is (-5, 3) and its radius is 6.

**7. x ^{2}Â + y^{2}Â â€“ 4x â€“ 8y â€“ 45 = 0**

**Solution:**

Given:

The equation of the given circle is x^{2}Â + y^{2}Â â€“ 4x â€“ 8y â€“ 45 = 0.

x^{2}Â + y^{2}Â â€“ 4x â€“ 8y â€“ 45 = 0

(x^{2}Â â€“ 4x) + (y^{2}Â -8y) = 45

(x^{2} â€“ 2(x) (2) + 2^{2}) + (y^{2} â€“ 2(y) (4) + 4^{2}) â€“ 4 â€“ 16 = 45

(x â€“ 2)^{2}Â + (y â€“ 4)^{2}Â = 65

(x â€“ 2)^{2}Â + (y â€“ 4)^{2}Â = (âˆš65)^{2} [which is form (x-h)^{2}Â +(y-k)^{2}Â = r^{2}]

Where h = 2, K = 4 and r = âˆš65

âˆ´ The centre of the given circle is (2, 4) and its radius is âˆš65.

**8. x ^{2}Â + y^{2}Â â€“ 8x + 10y â€“ 12 = 0**

**Solution:**

Given:

The equation of the given circle is x^{2}Â + y^{2}Â -8x + 10y -12 = 0.

x^{2}Â + y^{2}Â â€“ 8x + 10y â€“ 12 = 0

(x^{2}Â â€“ 8x) + (y^{2 }+ 10y) = 12

(x^{2} â€“ 2(x) (4) + 4^{2}) + (y^{2} â€“ 2(y) (5) + 5^{2}) â€“ 16 â€“ 25 = 12

(x â€“ 4)^{2}Â + (y + 5)^{2}Â = 53

(x â€“ 4)^{2}Â + (y â€“ (-5))^{2}Â = (âˆš53)^{2} [which is form (x-h)^{2}Â +(y-k)^{2}Â = r^{2}]

Where h = 4, K= -5 and r = âˆš53

âˆ´ The centre of the given circle is (4, -5) and its radius is âˆš53.

**9. 2x ^{2} + 2y^{2} â€“ x = 0**

**Solution:**

The equation of the given of the circle is 2x^{2}Â + 2y^{2}Â â€“x = 0.

2x^{2}Â + 2y^{2}Â â€“x = 0

(2x^{2}Â + x) + 2y^{2}Â = 0

(x^{2} â€“ 2 (x) (1/4) + (1/4)^{2}) + y^{2} â€“ (1/4)^{2} = 0

(x â€“ 1/4)^{2} + (y â€“ 0)^{2} = (1/4)^{2} [which is form (x-h)^{2}Â +(y-k)^{2}Â = r^{2}]

Where, h = Â¼, K = 0, and r = Â¼

âˆ´ The center of the given circle is (1/4, 0)Â and its radius isÂ 1/4.

**10. Find the equation of the circle passing through the points (4,1) and (6,5) and whose centre is on the line 4x + y = 16.**

**Solution:**

Let us consider the equation of the required circle be (x â€“ h)^{2}+ (y â€“ k)^{2}Â = r^{2}

We know that the circle passes through points (4,1) and (6,5)

So,

(4 â€“ h)^{2 }+ (1 â€“ k)^{2}Â = r^{2}Â â€¦â€¦â€¦â€¦â€¦..(1)

(6â€“ h)^{2}+ (5 â€“ k)^{2}Â = r^{2}Â â€¦â€¦â€¦â€¦â€¦â€¦(2)

Since, the centre (h, k) of the circle lies on line 4x + y = 16,

4h + k =16â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (3)

From the equation (1) and (2), we obtain

(4 â€“ h)^{2}+ (1 â€“ k)^{2}Â =(6 â€“ h)^{2}Â + (5 â€“ k)^{2}

16 â€“ 8h + h^{2}Â +1 -2k +k^{2}Â = 36 -12h +h^{2}+15 â€“ 10k + k^{2}

16 â€“ 8h +1 -2k + 12h -25 -10k

4h +8k = 44

h + 2k =11â€¦â€¦â€¦â€¦â€¦. (4)

On solving equations (3) and (4), we obtain h=3 and k= 4.

On substituting the values of h and k in equation (1), we obtain

(4 â€“ 3)^{2}+ (1 â€“ 4)^{2}Â = r^{2}

(1)^{2}Â + (-3)^{2}Â = r^{2}

1+9 = r^{2}

r =Â âˆš10

so now, (x â€“ 3)^{2 }+ (y â€“ 4)^{2}Â = (âˆš10)^{2}

x^{2}Â â€“ 6x + 9 + y^{2}Â â€“ 8y + 16 =10

x^{2}Â + y^{2}Â â€“ 6x â€“ 8y + 15 = 0

âˆ´ The equation of the required circle is x^{2}Â + y^{2}Â â€“ 6x â€“ 8y + 15 = 0

**11. Find the equation of the circle passing through the points (2, 3) and (â€“1, 1) and whose centre is on the line x â€“ 3y â€“ 11 = 0.**

**Solution: **

Let us consider the equation of the required circle be (x â€“ h)^{2 }+ (y â€“ k)^{2}Â = r^{2}

We know that the circle passes through points (2,3) and (-1,1).

(2 â€“ h)^{2}+ (3 â€“ k)^{2}Â =r^{2}Â â€¦â€¦â€¦â€¦â€¦..(1)

(-1 â€“ h)^{2}+ (1â€“ k)^{2}Â =r^{2}Â â€¦â€¦â€¦â€¦â€¦â€¦(2)

Since, the centre (h, k) of the circle lies on line x â€“ 3y â€“ 11= 0,

h â€“ 3k =11â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (3)

From the equation (1) and (2), we obtain

(2 â€“ h)^{2}+ (3 â€“ k)^{2}Â =(-1 â€“ h)^{2}Â + (1 â€“ k)^{2}

4 â€“ 4h + h^{2}Â +9 -6k +k^{2}Â = 1 + 2h +h^{2}+1 â€“ 2k + k^{2}

4 â€“ 4h +9 -6k = 1 + 2h + 1 -2k

6h + 4k =11â€¦â€¦â€¦â€¦â€¦. (4)

Now let us multiply equation (3) by 6 and subtract it from equation (4) to get,

6h+ 4k â€“ 6(h-3k) = 11 â€“ 66

6h + 4k â€“ 6h + 18k = 11 â€“ 66

22 k = â€“ 55

K = -5/2

Substitute this value of K in equation (4) to get,

6h + 4(-5/2) = 11

6h â€“ 10 = 11

6h = 21

h = 21/6

h = 7/2

We obtain h =Â 7/2and k = -5/2

On substituting the values of h and k in equation (1), we get

(2 â€“ 7/2)^{2} + (3 + 5/2)^{2} = r^{2}

^{2}+ [(6+5)/2]

^{2}= r

^{2}

(-3/2)^{2} + (11/2)^{2} = r^{2}

9/4 + 121/4 = r^{2}

130/4 = r^{2}

The equation of the required circle is

(x â€“ 7/2)^{2} + (y + 5/2)^{2} = 130/4

^{2}+ [(2y+5)/2]

^{2}= 130/4

4x^{2}Â -28x + 49 +4y^{2}Â + 20y + 25 =130

4x^{2}Â +4y^{2}Â -28x + 20y â€“ 56 = 0

4(x^{2}Â +y^{2}Â -7x + 5y â€“ 14) = 0

x^{2 }+ y^{2 }â€“ 7x + 5y â€“ 14 = 0

âˆ´ The equation of the required circle is x^{2 }+ y^{2 }â€“ 7x + 5y â€“ 14 = 0

**12. Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).**

**Solution:**

Let us consider the equation of the required circle be (x â€“ h)^{2}+ (y â€“ k)^{2}Â = r^{2}

We know that the radius of the circle is 5 and its centre lies on the x-axis, k = 0 and r = 5.

So now, the equation of the circle is (x â€“ h)^{2}Â + y^{2}Â = 25.

It is given that the circle passes through the point (2, 3) so the point will satisfy the equation of the circle.

(2 â€“ h)^{2}+ 3^{2}Â = 25

(2 â€“ h)^{2}Â = 25-9

(2 â€“ h)^{2}Â = 16

2 â€“ h = **Â±** **âˆš**16 = **Â±** 4

If 2-h = 4, then h = -2

If 2-h = -4, then h = 6

Then, when h = -2, the equation of the circle becomes

(x + 2)^{2}Â + y^{2}Â = 25

x^{2}Â + 12x + 36 + y^{2}Â = 25

x^{2}Â + y^{2}Â + 4x â€“ 21 = 0

When h = 6, the equation of the circle becomes

(x â€“ 6)^{2}Â + y^{2}Â = 25

x^{2}Â -12x + 36 + y^{2}Â = 25

x^{2}Â + y^{2}Â -12x + 11 = 0

âˆ´ The equation of the required circle is x^{2}Â + y^{2}Â + 4x â€“ 21 = 0 and x^{2}Â + y^{2}Â -12x + 11 = 0

**13. Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.**

**Solution:**

Let us consider the equation of the required circle be (x â€“ h)^{2}+ (y â€“ k)^{2}Â =r^{2}

We know that the circle passes through (0, 0),

So, (0 â€“ h)^{2}+ (0 â€“ k)^{2}Â = r^{2}

h^{2}Â + k^{2}Â =Â r^{2}

Now, The equation of the circle is (x â€“ h)^{2 }+ (y â€“ k)^{2}Â = h^{2}Â + k^{2}.

It is given that the circle intercepts a and b on the coordinate axes.

i.e., the circle passes through points (a, 0) and (0, b).

So, (a â€“ h)^{2}+ (0 â€“ k)^{2}Â =h^{2}Â +k^{2}â€¦â€¦â€¦â€¦â€¦..(1)

(0 â€“ h)^{2}+ (bâ€“ k)^{2}Â =h^{2}Â +k^{2}â€¦â€¦â€¦â€¦â€¦â€¦(2)

From equation (1), we obtain

a^{2}Â â€“ 2ah + h^{2}Â +k^{2}Â = h^{2}Â +k^{2}

a^{2}Â â€“ 2ah = 0

a(a â€“ 2h) =0

a = 0 or (a -2h) = 0

However, a â‰ Â 0; hence, (a -2h) = 0

h = a/2

From equation (2), we obtain

h^{2}Â â€“ 2bk + k^{2}Â + b^{2}= h^{2}Â +k^{2}

b^{2}Â â€“ 2bk = 0

b(bâ€“ 2k) = 0

b= 0 or (b-2k) =0

However, aÂ â‰ 0; hence, (b -2k) = 0

kÂ =b/2

So, the equation is

(x â€“ a/2)^{2} + (y â€“ b/2)^{2} = (a/2)^{2} + (b/2)^{2}

^{2}+ [(2y-b)/2]

^{2}= (a

^{2}+ b

^{2})/4

4x^{2}Â â€“ 4ax + a^{2}Â +4y^{2}Â â€“ 4by + b^{2}Â = a^{2}Â + b^{2}

4x^{2}Â + 4y^{2}Â -4ax â€“ 4by = 0

4(x^{2}Â +y^{2}Â -7x + 5y â€“ 14) = 0

x^{2 }+ y^{2 }â€“ axÂ â€“ by = 0

âˆ´ The equation of the required circle is x^{2 }+ y^{2 }â€“ axÂ â€“ by = 0

**14. Find the equation of a circle with centre (2,2) and passes through the point (4,5).**

**Solution:**

Given:

The centre of the circle is given as (h, k) = (2,2)

We know that the circle passes through point (4,5), the radius (r) of the circle is the distance between the points (2,2) and (4,5).

r = **âˆš**[(2-4)^{2} + (2-5)^{2}]

= âˆš[(-2)^{2} + (-3)^{2}]

= âˆš[4+9]

= âˆš13

The equation of the circle is given as

(xâ€“ h)^{2}+ (y â€“ k)^{2}Â = r^{2}

(x â€“h)^{2}Â + (y â€“ k)^{2}Â = (âˆš13)^{2}

(x â€“2)^{2}Â + (y â€“ 2)^{2}Â = (âˆš13)^{2}

x^{2}Â â€“ 4x + 4 + y^{2 }â€“ 4y + 4 = 13

x^{2}Â + y^{2}Â â€“ 4x â€“ 4y = 5

âˆ´ The equation of the required circle is x^{2}Â + y^{2}Â â€“ 4x â€“ 4y = 5

**15. Does the point (â€“2.5, 3.5) lie inside, outside or on the circle x ^{2}Â + y^{2}Â = 25?**

**Solution:**

Given:

The equation of the given circle isÂ x^{2}Â +y^{2}Â = 25.

x^{2}Â + y^{2}Â = 25

(x â€“ 0)^{2}Â + (y â€“ 0)^{2}Â = 5^{2} [which is of the form (x â€“ h)^{2 }+ (y â€“ k)^{2}Â = r^{2}]

Where, h = 0, k = 0 and r = 5.

So the distance between point (-2.5, 3.5) and the centre (0,0) is

= âˆš[(-2.5 â€“ 0)^{2} + (-3.5 â€“ 0)^{2}]

= âˆš(6.25 + 12.25)

= âˆš18.5

= 4.3 [which is < 5]

Since, the distance between point (-2.5, -3.5) and the centre (0, 0) of the circle is less than the radius of the circle, point (-2.5, -3.5) lies inside the circle.

EXERCISE 11.2 PAGE NO: 246

**In each of the following Exercises 1 to 6, find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum.**

1. y^{2}Â = 12x

**Solution:**

Given:

The equation is y^{2}Â = 12x

Here we know that the coefficient of x is positive.

So, the parabola opens towards the right.

On comparing this equation with y^{2}Â = 4ax, we get,

4a = 12

a = 3

Thus, the co-ordinates of the focus = (a, 0) = (3, 0)

Since, the given equation involves y^{2}, the axis of the parabola is the x-axis.

âˆ´ The equation of directrix, x = -a, then,

x + 3 = 0

Length of latus rectum = 4a = 4 Ã— 3 = 12

**2. x ^{2}Â = 6y**

**Solution:**

Given:

The equation is x^{2}Â = 6y

Here we know that the coefficient of y is positive.

So, the parabola opens upwards.

On comparing this equation with x^{2}Â = 4ay, we get,

4a = 6

a = 6/4

= 3/2

Thus, the co-ordinates of the focus = (0,a) = (0, 3/2)

Since, the given equation involves x^{2}, the axis of the parabola is the y-axis.

âˆ´ The equation of directrix, y =-a, then,

y = -3/2

Length of latus rectum = 4a = 4(3/2) = 6

**3. y ^{2}Â = â€“ 8x**

**Solution:**

Given:

The equation is y^{2}Â = -8x

Here we know that the coefficient of x is negative.

So, the parabola open towards the left.

On comparing this equation with y^{2}Â = -4ax, we get,

-4a = -8

a = -8/-4 = 2

Thus, co-ordinates of the focus = (-a,0) = (-2, 0)

Since, the given equation involves y^{2}, the axis of the parabola is the x-axis.

âˆ´ Equation of directrix, x =a, then,

x = 2

Length of latus rectum = 4a = 4 (2) = 8

**4. x ^{2}Â = â€“ 16y**

**Solution:**

Given:

The equation is x^{2}Â = -16y

Here we know that the coefficient of y is negative.

So, the parabola opens downwards.

On comparing this equation with x^{2}Â = -4ay, we get,

-4a = -16

a = -16/-4

= 4

Thus, co-ordinates of the focus = (0,-a) = (0,-4)

Since, the given equation involves x^{2}, the axis of the parabola is the y-axis.

âˆ´ The equation of directrix, y =a, then,

y = 4

Length of latus rectum = 4a = 4(4) = 16

**5. y ^{2}Â = 10x**

**Solution:**

Given:

The equation is y^{2}Â = 10x

Here we know that the coefficient of x is positive.

So, the parabola open towards the right.

On comparing this equation with y^{2}Â = 4ax, we get,

4a = 10

a =Â 10/4 = 5/2

Thus, co-ordinates of the focus = (a,0) = (5/2, 0)

Since, the given equation involves y^{2}, the axis of the parabola is the x-axis.

âˆ´ The equation of directrix, x =-a, then,

x = â€“ 5/2

Length of latus rectum = 4a = 4(5/2) = 10

**6. x ^{2}Â = â€“ 9y**

**Solution:**

Given:

The equation is x^{2}Â = -9y

Here we know that the coefficient of y is negative.

So, the parabola open downwards.

On comparing this equation with x^{2}Â = -4ay, we get,

-4a = -9

a = -9/-4 = 9/4

Thus, co-ordinates of the focus = (0,-a) = (0, -9/4)

Since, the given equation involves x^{2}, the axis of the parabola is the y-axis.

âˆ´ The equation of directrix, y = a, then,

y =Â 9/4

Length of latus rectum = 4a = 4(9/4) = 9

**In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions: **

7. Focus (6,0); directrix x = â€“ 6

**Solution:**

Given:

Focus (6,0) and directrix x = -6

We know that the focus lies on the xâ€“axis is the axis of the parabola.

So, the equation of the parabola is either of the form y^{2}Â = 4ax or y^{2}Â = -4ax.

It is also seen that the directrix, x = -6 is to the left of the y- axis,

While the focus (6, 0) is to the right of the y â€“axis.

Hence, the parabola is of the form y^{2}Â = 4ax.

Here, a = 6

âˆ´ The equation of the parabola is y^{2}Â = 24x.

**8. Focus (0,â€“3); directrix y = 3**

**Solution:**

Given:

Focus (0, -3) and directrix y = 3

We know that the focus lies on the yâ€“axis, the y-axis is the axis of the parabola.

So, the equation of the parabola is either of the form x^{2}Â = 4ay or x^{2}Â = -4ay.

It is also seen that the directrix, y = 3 is above the x- axis,

While the focus (0,-3) is below the x-axis.

Hence, the parabola is of the form x^{2}Â = -4ay.

Here, a = 3

âˆ´ The equation of the parabola is x^{2}Â = -12y.

**9. Vertex (0, 0); focus (3, 0)**

**Solution:**

Given:

Vertex (0, 0) and focus (3, 0)

We know that the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis. [x-axis is the axis of the parabola.]

The equation of the parabola is of the form y^{2 }= 4ax.

Since, the focus is (3, 0), a = 3

âˆ´ The equation of the parabola is y^{2}Â = 4 Ã— 3 Ã— x,

y^{2}Â = 12x

**10. Vertex (0, 0); focus (â€“2, 0)**

**Solution:**

Given:

Vertex (0, 0) and focus (-2, 0)

We know that the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis. [x-axis is the axis of the parabola.]

The equation of the parabola is of the form y^{2}=-4ax.

Since, the focus is (-2, 0), a = 2

âˆ´ The equation of the parabola is y^{2}Â = -4 Ã— 2 Ã— x,

y^{2}Â = -8x

**11. Vertex (0, 0) passing through (2, 3) and axis is along x-axis.**

**Solution:**

We know that the vertex is (0, 0) and the axis of the parabola is the x-axis

The equation of the parabola is either of the from y^{2 }= 4ax or y^{2}Â = -4ax.

Given that the parabola passes through point (2, 3), which lies in the first quadrant.

So, the equation of the parabola is of the form y^{2}Â = 4ax, while point (2, 3) must satisfy the equation y^{2}Â = 4ax.

Then,

3^{2}Â = 4a(2)

3^{2} = 8a

9 = 8a

a =Â 9/8

Thus, the equation of the parabola is

y^{2} = 4 (9/8)x

= 9x/2

2y^{2} = 9x

âˆ´ The equation of the parabola is 2y^{2} = 9x

**12. Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.**

**Solution:**

We know that the vertex is (0, 0) and the parabola is symmetric about the y-axis.

The equation of the parabola is either of the from x^{2 }= 4ay or x^{2}Â = -4ay.

Given that the parabola passes through point (5, 2), which lies in the first quadrant.

So, the equation of the parabola is of the form x^{2}Â = 4ay, while point (5, 2) must satisfy the equation x^{2}Â = 4ay.

Then,

5^{2}Â = 4a(2)

25 = 8a

a =Â 25/8

Thus, the equation of the parabola is

x^{2} = 4 (25/8)y

x^{2} = 25y/2

2x^{2} = 25y

âˆ´ The equation of the parabola is 2x^{2} = 25y

EXERCISE 11.3 PAGE NO: 255

**In each of the Exercises 1 to 9, find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.**

**1. x ^{2}/36 + y^{2}/16 = 1**

**Solution:**

Given:

The equation is x^{2}/36 + y^{2}/16 = 1

Here, the denominator of x^{2}/36Â is greater than the denominator of y^{2}/16.

So, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation withÂ x^{2}/a^{2} + y^{2}/b^{2} = 1, we get

a = 6 and b = 4.

c = âˆš(a^{2}Â â€“ b^{2})

= âˆš(36-16)

= âˆš20

= 2âˆš5

Then,

The coordinates of the foci are (2âˆš5, 0) and (-2âˆš5, 0).

The coordinates of the vertices are (6, 0) and (-6, 0)

Length of major axis = 2a = 2 (6) = 12

Length of minor axis = 2b = 2 (4) = 8

Eccentricity, e = c/aÂ = 2âˆš5/6 = âˆš5/3

Length of latus rectum = 2b^{2}/a = (2Ã—16)/6 = 16/3

**2. x ^{2}/4 + y^{2}/25 = 1**

**Solution:**

Given:

The equation is x^{2}/4 + y^{2}/25 = 1

Here, the denominator of y^{2}/25Â is greater than the denominator of x^{2}/4.

So, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation with x^{2}/a^{2} + y^{2}/b^{2} = 1, we get

a = 5 and b = 2.

c = âˆš(a^{2}Â â€“ b^{2})

= âˆš(25-4)

= âˆš21

Then,

The coordinates of the foci are (0, âˆš21) and (0, -âˆš21).

The coordinates of the vertices are (0, 5) and (0, -5)

Length of major axis = 2a = 2 (5) = 10

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/aÂ = âˆš21/5

Length of latus rectum = 2b^{2}/a = (2Ã—2^{2})/5 = (2Ã—4)/5 = 8/5

**3. Â x ^{2}/16 + y^{2}/9 = 1**

**Solution:**

Given:

The equation is x^{2}/16 + y^{2}/9 = 1 or x^{2}/4^{2} + y^{2}/3^{2} = 1

Here, the denominator of x^{2}/16Â is greater than the denominator of y^{2}/9.

So, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation withÂ x^{2}/a^{2} + y^{2}/b^{2} = 1, we get

a = 4 and b = 3.

c = âˆš(a^{2}Â â€“ b^{2})

= âˆš(16-9)

= âˆš7

Then,

The coordinates of the foci are (âˆš7, 0) and (-âˆš7, 0).

The coordinates of the vertices are (4, 0) and (-4, 0)

Length of major axis = 2a = 2 (4) = 8

Length of minor axis = 2b = 2 (3) = 6

Eccentricity, e = c/aÂ = âˆš7/4

Length of latus rectum = 2b^{2}/a = (2Ã—3^{2})/4 = (2Ã—9)/4 = 18/4 = 9/2

**4. Â x ^{2}/25 + y^{2}/100 = 1**

**Solution:**

Given:

The equation is x^{2}/25 + y^{2}/100 = 1

Here, the denominator of y^{2}/100Â is greater than the denominator of x^{2}/25.

So, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation withÂ x^{2}/b^{2} + y^{2}/a^{2} = 1, we get

b = 5 and a =10.

c = âˆš(a^{2}Â â€“ b^{2})

= âˆš(100-25)

= âˆš75

= 5âˆš3

Then,

The coordinates of the foci are (0, 5âˆš3) and (0, -5âˆš3).

The coordinates of the vertices are (0, âˆš10) and (0, -âˆš10)

Length of major axis = 2a = 2 (10) = 20

Length of minor axis = 2b = 2 (5) = 10

Eccentricity, e = c/aÂ = 5âˆš3/10 = âˆš3/2

Length of latus rectum = 2b^{2}/a = (2Ã—5^{2})/10 = (2Ã—25)/10 = 5

**5. x ^{2}/49 + y^{2}/36 = 1**

**Solution:**

Given:

The equation is x^{2}/49 + y^{2}/36 = 1

Here, the denominator of x^{2}/49Â is greater than the denominator of y^{2}/36.

So, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation withÂ x^{2}/a^{2} + y^{2}/b^{2} = 1, we get

b = 6 and a =7

c = âˆš(a^{2}Â â€“ b^{2})

= âˆš(49-36)

= âˆš13

Then,

The coordinates of the foci are (âˆš13, 0) and (-âˆš3, 0).

The coordinates of the vertices are (7, 0) and (-7, 0)

Length of major axis = 2a = 2 (7) = 14

Length of minor axis = 2b = 2 (6) = 12

Eccentricity, e = c/aÂ = âˆš13/7

Length of latus rectum = 2b^{2}/a = (2Ã—6^{2})/7 = (2Ã—36)/7 = 72/7

**6. x ^{2}/100 + y^{2}/400 = 1**

**Solution:**

Given:

The equation is x^{2}/100 + y^{2}/400 = 1

Here, the denominator of y^{2}/400Â is greater than the denominator of x^{2}/100.

So, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation withÂ x^{2}/b^{2} + y^{2}/a^{2} = 1, we get

b = 10 and a =20.

c = âˆš(a^{2}Â â€“ b^{2})

= âˆš(400-100)

= âˆš300

= 10âˆš3

Then,

The coordinates of the foci are (0, 10âˆš3) and (0, -10âˆš3).

The coordinates of the vertices are (0, 20) and (0, -20)

Length of major axis = 2a = 2 (20) = 40

Length of minor axis = 2b = 2 (10) = 20

Eccentricity, e = c/aÂ = 10âˆš3/20 = âˆš3/2

Length of latus rectum = 2b^{2}/a = (2Ã—10^{2})/20 = (2Ã—100)/20 = 10

**7. 36x ^{2}Â + 4y^{2}Â = 144**

**Solution:**

Given:

The equation is 36x^{2}Â + 4y^{2}Â = 144 or x^{2}/4 + y^{2}/36 = 1 or x^{2}/2^{2} + y^{2}/6^{2} = 1

Here, the denominator of y^{2}/6^{2}Â is greater than the denominator of x^{2}/2^{2}.

So, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation withÂ x^{2}/b^{2} + y^{2}/a^{2} = 1, we get

b = 2 and a = 6.

c = âˆš(a^{2}Â â€“ b^{2})

= âˆš(36-4)

= âˆš32

= 4âˆš2

Then,

The coordinates of the foci are (0, 4âˆš2) and (0, -4âˆš2).

The coordinates of the vertices are (0, 6) and (0, -6)

Length of major axis = 2a = 2 (6) = 12

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/aÂ = 4âˆš2/6 = 2âˆš2/3

Length of latus rectum = 2b^{2}/a = (2Ã—2^{2})/6 = (2Ã—4)/6 = 4/3

**8. 16x ^{2}Â + y^{2}Â = 16**

**Solution:**

Given:

The equation is 16x^{2}Â + y^{2}Â = 16 or x^{2}/1 + y^{2}/16 = 1 or x^{2}/1^{2} + y^{2}/4^{2} = 1

Here, the denominator of y^{2}/4^{2}Â is greater than the denominator of x^{2}/1^{2}.

So, the major axis is along the y-axis, while the minor axis is along the x-axis.

On comparing the given equation withÂ x^{2}/b^{2} + y^{2}/a^{2} = 1, we get

b =1 and a =4.

c = âˆš(a^{2}Â â€“ b^{2})

= âˆš(16-1)

= âˆš15

Then,

The coordinates of the foci are (0, âˆš15) and (0, -âˆš15).

The coordinates of the vertices are (0, 4) and (0, -4)

Length of major axis = 2a = 2 (4) = 8

Length of minor axis = 2b = 2 (1) = 2

Eccentricity, e = c/aÂ = âˆš15/4

Length of latus rectum = 2b^{2}/a = (2Ã—1^{2})/4 = 2/4 = Â½

**9. 4x ^{2}Â + 9y^{2}Â = 36**

**Solution:**

Given:

The equation is 4x^{2}Â + 9y^{2}Â = 36 or x^{2}/9 + y^{2}/4 = 1 or x^{2}/3^{2} + y^{2}/2^{2} = 1

Here, the denominator of x^{2}/3^{2}Â is greater than the denominator of y^{2}/2^{2}.

So, the major axis is along the x-axis, while the minor axis is along the y-axis.

On comparing the given equation withÂ x^{2}/a^{2} + y^{2}/b^{2} = 1, we get

a =3 and b =2.

c = âˆš(a^{2}Â â€“ b^{2})

= âˆš(9-4)

= âˆš5

Then,

The coordinates of the foci are (âˆš5, 0) and (-âˆš5, 0).

The coordinates of the vertices are (3, 0) and (-3, 0)

Length of major axis = 2a = 2 (3) = 6

Length of minor axis = 2b = 2 (2) = 4

Eccentricity, e = c/aÂ = âˆš5/3

Length of latus rectum = 2b^{2}/a = (2Ã—2^{2})/3 = (2Ã—4)/3 = 8/3

**In each of the following Exercises 10 to 20, find the equation for the ellipse that satisfies the given conditions:**

**10. Vertices (Â±Â 5, 0), foci (Â±Â 4, 0)**

**Solution:**

Given:

Vertices (Â±Â 5, 0) and foci (Â±Â 4, 0)

Here, the vertices are on the x-axis.

So, the equation of the ellipse will be of the formÂ x^{2}/a^{2} + y^{2}/b^{2} = 1, where â€˜aâ€™ is the semi-major axis.

Then, a = 5 and c = 4.

It is known that a^{2}Â = b^{2 }+ c^{2}.

So, 5^{2}Â = b^{2 }+ 4^{2}

25 = b^{2}Â + 16

b^{2}Â = 25 â€“ 16

b =Â âˆš9

= 3

âˆ´ The equation of the ellipse is x^{2}/5^{2} + y^{2}/3^{2} = 1 or x^{2}/25 + y^{2}/9Â = 1

**11. Vertices (0,Â Â±Â 13), foci (0,Â Â± 5)**

**Solution:**

Given:

Vertices (0,Â Â±Â 13) and foci (0,Â Â± 5)

Here, the vertices are on the y-axis.

So, the equation of the ellipse will be of the formÂ x^{2}/b^{2} + y^{2}/a^{2} = 1, where â€˜aâ€™ is the semi-major axis.

Then, a =13 and c = 5.

It is known that a^{2}Â = b^{2 }+ c^{2}.

13^{2}Â = b^{2}+5^{2}

169 = b^{2}Â + 15

b^{2}Â = 169 â€“ 125

b =Â âˆš144

= 12

âˆ´ The equation of the ellipse is x^{2}/12^{2} + y^{2}/13^{2}Â = 1 or x^{2}/144 + y^{2}/169Â = 1

**12. Vertices (Â±Â 6, 0), foci (Â± 4, 0)**

**Solution:**

Given:

Vertices (Â±Â 6, 0) and foci (Â± 4, 0)

Here, the vertices are on the x-axis.

So, the equation of the ellipse will be of the formÂ x^{2}/a^{2} + y^{2}/b^{2} = 1, where â€˜aâ€™ is the semi-major axis.

Then, a = 6 and c = 4.

It is known that a^{2}Â = b^{2 }+ c^{2}.

6^{2}Â = b^{2}+4^{2}

36 = b^{2}Â + 16

b^{2}Â = 36 â€“ 16

b =Â âˆš20

âˆ´ The equation of the ellipse is x^{2}/6^{2} + y^{2}/(âˆš20)^{2}Â = 1 or x^{2}/36 + y^{2}/20Â = 1

**13. Ends of major axis (Â± 3, 0), ends of minor axis (0,Â Â±2)**

**Solution:**

Given:

Ends of major axis (Â± 3, 0) and ends of minor axis (0,Â Â±2)

Here, the major axis is along the x-axis.

So, the equation of the ellipse will be of the formÂ x^{2}/a^{2} + y^{2}/b^{2} = 1, where â€˜aâ€™ is the semi-major axis.

Then, a = 3 and b = 2.

âˆ´ The equation for the ellipse x^{2}/3^{2} + y^{2}/2^{2}Â = 1 or x^{2}/9 + y^{2}/4Â = 1

**14. Ends of major axis (0,Â Â±âˆš5), ends of minor axis (Â±1, 0)**

**Solution:**

Given:

Ends of major axis (0,Â Â±âˆš5) and ends of minor axis (Â±1, 0)

Here, the major axis is along the y-axis.

So, the equation of the ellipse will be of the formÂ x^{2}/b^{2} + y^{2}/a^{2} = 1, where â€˜aâ€™ is the semi-major axis.

Then, a =Â âˆš5 and b = 1.

âˆ´ The equation for the ellipse x^{2}/1^{2} + y^{2}/(âˆš5)^{2}Â = 1 or x^{2}/1 + y^{2}/5Â = 1

**15. Length of major axis 26, foci (Â±5, 0)**

**Solution:**

Given:

Length of major axis is 26 and foci (Â±5, 0)

Since the foci are on the x-axis, the major axis is along the x-axis.

^{2}/a^{2} + y^{2}/b^{2} = 1, where â€˜aâ€™ is the semi-major axis.

Then, 2a = 26

a = 13 and c = 5.

It is known that a^{2}Â = b^{2 }+ c^{2}.

13^{2}Â = b^{2}+5^{2}

169 = b^{2}Â + 25

b^{2}Â = 169 â€“ 25

b =Â âˆš144

= 12

âˆ´ The equation of the ellipse is x^{2}/13^{2} + y^{2}/12^{2} = 1Â or x^{2}/169 + y^{2}/144Â = 1

**16. Length of minor axis 16, foci (0,Â Â±6).**

**Solution:**

Given:

Length of minor axis is 16 and foci (0,Â Â±6).

Since the foci are on the y-axis, the major axis is along the y-axis.

So, the equation of the ellipse will be of the formÂ x^{2}/b^{2} + y^{2}/a^{2} = 1, where â€˜aâ€™ is the semi-major axis.

Then, 2b =16

b = 8 and c = 6.

It is known that a^{2}Â = b^{2 }+ c^{2}.

a^{2}Â = 8^{2 }+ 6^{2}

=Â 64 + 36

=100

a =Â âˆš100

= 10

âˆ´ The equation of the ellipse is x^{2}/8^{2} + y^{2}/10^{2}Â =1 or x^{2}/64 + y^{2}/100Â = 1

**17. Foci (Â±3, 0), a = 4**

**Solution:**

Given:

Foci (Â±3, 0) and a = 4

Since the foci are on the x-axis, the major axis is along the x-axis.

So, the equation of the ellipse will be of the form x^{2}/a^{2} + y^{2}/b^{2} = 1, where â€˜aâ€™ is the semi-major axis.

Then, c = 3 and a = 4.

It is known that a^{2}Â = b^{2 }+ c^{2}.

a^{2}Â = 8^{2 }+ 6^{2}

= 64 + 36

= 100

16 = b^{2}Â + 9

b^{2}Â = 16 â€“ 9

= 7

âˆ´ The equation of the ellipse is x^{2}/16 + y^{2}/7 = 1

**18. b = 3, c = 4, centre at the origin; foci on the x axis.**

**Solution:**

Given:

b = 3, c = 4, centre at the origin and foci on the x axis.

Since the foci are on the x-axis, the major axis is along the x-axis.

^{2}/a^{2} + y^{2}/b^{2} = 1, where â€˜aâ€™ is the semi-major axis.

Then, b = 3 and c = 4.

It is known that a^{2}Â = b^{2 }+ c^{2}.

a^{2}Â = 3^{2 }+ 4^{2}

= 9 + 16

=25

a = âˆš25

= 5

âˆ´ The equation of the ellipse is x^{2}/5^{2} + y^{2}/3^{2} or x^{2}/25 + y^{2}/9 = 1

**19. Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).**

**Solution:**

Given:

Centre at (0, 0), major axis on the y-axis and passes through the points (3, 2) and (1, 6).

Since the centre is at (0, 0) and the major axis is on the y- axis, the equation of the ellipse will be of the form x^{2}/b^{2} + y^{2}/a^{2} = 1, where â€˜aâ€™ is the semi-major axis.

The ellipse passes through points (3, 2) and (1, 6).

So, by putting the values x = 3 and y = 2, we get,

3^{2}/b^{2} + 2^{2}/a^{2} = 1

9/b^{2} + 4/a^{2}â€¦. (1)

And by putting the values x = 1 and y = 6, we get,

1^{1}/b^{2} + 6^{2}/a^{2} = 1

1/b^{2} + 36/a^{2} = 1 â€¦. (2)

On solving equation (1) and (2), we get

b^{2}Â = 10 and a^{2}Â = 40.

âˆ´ The equation of the ellipse is x^{2}/10 + y^{2}/40Â = 1 or 4x^{2}Â + yÂ ^{2}Â = 40

**20. Major axis on the x-axis and passes through the points (4,3) and (6,2).**

**Solution:**

Given:

Major axis on the x-axis and passes through the points (4, 3) and (6, 2).

Since the major axis is on the x-axis, the equation of the ellipse will be the form

x^{2}/a^{2} + y^{2}/b^{2} = 1â€¦. (1) [Where â€˜aâ€™ is the semi-major axis.]

The ellipse passes through points (4, 3) and (6, 2).

So by putting the values x = 4 and y = 3 in equation (1), we get,

16/a^{2} + 9/b^{2} = 1 â€¦. (2)

Putting, x = 6 and y = 2 in equation (1), we get,

36/a^{2} + 4/b^{2} = 1 â€¦. (3)

From equation (2)

16/a^{2} = 1 â€“ 9/b^{2}

1/a^{2} = (1/16 (1 â€“ 9/b^{2})) â€¦. (4)

Substituting the value of 1/a^{2} in equation (3) we get,

36/a^{2} + 4/b^{2} = 1

36(1/a^{2}) + 4/b^{2} = 1

36[1/16 (1 â€“ 9/b^{2})] + 4/b^{2} = 1

36/16 (1 â€“ 9/b^{2}) + 4/b^{2} = 1

9/4 (1 â€“ 9/b^{2}) + 4/b^{2} = 1

9/4 â€“ 81/4b^{2} + 4/b^{2} = 1

-81/4b^{2} + 4/b^{2} = 1 â€“ 9/4

(-81+16)/4b^{2} = (4-9)/4

-65/4b^{2} = -5/4

-5/4(13/b^{2}) = -5/4

13/b^{2} = 1

1/b^{2} = 1/13

b^{2} = 13

Now substitute the value of b^{2} in equation (4) we get,

1/a^{2} = 1/16(1 â€“ 9/b^{2})

= 1/16(1 â€“ 9/13)

= 1/16((13-9)/13)

= 1/16(4/13)

= 1/52

a^{2} = 52

Equation of ellipse is x^{2}/a^{2} + y^{2}/b^{2} = 1

By substituting the values of a^{2} and b^{2} in above equation we get,

x^{2}/52 + y^{2}/13 = 1

EXERCISE 11.4 PAGE NO: 262

**In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.**

**1. x ^{2}/16 â€“ y^{2}/9 = 1**

**Solution:**

Given:

The equation is x^{2}/16 â€“ y^{2}/9 = 1 or x^{2}/4^{2} â€“ y^{2}/3^{2} = 1

On comparing this equation with the standard equation of hyperbola x^{2}/a^{2} â€“ y^{2}/b^{2} = 1,

We get a = 4 and b = 3,

It is known that, a^{2}Â + b^{2}Â = c^{2}

So,

c^{2}Â = 4^{2}Â + 3^{2}

=** âˆš**25

c = 5

Then,

The coordinates of the foci are (**Â±**5, 0).

The coordinates of the vertices are (**Â±**4, 0).

Eccentricity, e = c/a = 5/4

Length of latus rectum = 2b^{2}/a = (2 Ã— 3^{2})/4 = (2Ã—9)/4 = 18/4 = 9/2

**2. y ^{2}/9 â€“ x^{2}/27 = 1**

**Solution:**

Given:

The equation is y^{2}/9 â€“ x^{2}/27 = 1 or y^{2}/3^{2} â€“ x^{2}/27^{2} = 1

On comparing this equation with the standard equation of hyperbola y^{2}/a^{2} â€“ x^{2}/b^{2} = 1,

We get a = 3 and b = **âˆš**27,

It is known that, a^{2}Â + b^{2}Â = c^{2}

So,

c^{2}Â = 3^{2}Â + (**âˆš**27)^{2}

= 9 + 27

c^{2} = 36

c = **âˆš**36

= 6

Then,

The coordinates of the foci are (0, 6) and (0, -6).

The coordinates of the vertices are (0, 3) and (0, â€“ 3).

Eccentricity, e = c/a = 6/3 = 2

Length of latus rectum = 2b^{2}/a = (2 Ã— 27)/3 = (54)/3 = 18

**3. 9y ^{2}Â â€“ 4x^{2}Â = 36**

**Solution:**

Given:

The equation is 9y^{2}Â â€“ 4x^{2}Â = 36 or y^{2}/4 â€“ x^{2}/9 = 1 or y^{2}/2^{2} â€“ x^{2}/3^{2} = 1

On comparing this equation with the standard equation of hyperbola y^{2}/a^{2} â€“ x^{2}/b^{2} = 1,

We get a = 2 and b = 3,

It is known that, a^{2}Â + b^{2}Â = c^{2}

So,

c^{2}Â = 4Â + 9

c^{2} = 13

c = **âˆš**13

Then,

The coordinates of the foci are (0, **âˆš**13) and (0, â€“**âˆš**13).

The coordinates of the vertices are (0, 2) and (0, â€“ 2).

Eccentricity, e = c/a = **âˆš**13/2

Length of latus rectum = 2b^{2}/a = (2 Ã— 3^{2})/2 = (2Ã—9)/2 = 18/2 = 9

**4. 16x ^{2}Â â€“ 9y^{2}Â = 576**

**Solution:**

Given:

The equation is 16x^{2}Â â€“ 9y^{2}Â = 576

Let us divide the whole equation by 576, we get

16x^{2}/576 â€“ 9y^{2}/576 = 576/576

x^{2}/36 â€“ y^{2}/64 = 1

On comparing this equation with the standard equation of hyperbola x^{2}/a^{2} â€“ y^{2}/b^{2} = 1,

We get a = 6 and b = 8,

It is known that, a^{2}Â + b^{2}Â = c^{2}

So,

c^{2}Â = 36Â + 64

c^{2} = **âˆš**100

c = 10

Then,

The coordinates of the foci are (10, 0) and (-10, 0).

The coordinates of the vertices are (6, 0) and (-6, 0).

Eccentricity, e = c/a = 10/6 = 5/3

Length of latus rectum = 2b^{2}/a = (2 Ã— 8^{2})/6 = (2Ã—64)/6 = 64/3

**5. 5y ^{2}Â â€“ 9x^{2}Â = 36**

**Solution:**

Given:

The equation is 5y^{2}Â â€“ 9x^{2}Â = 36

Let us divide the whole equation by 36, we get

5y^{2}/36 â€“ 9x^{2}/36 = 36/36

y^{2}/(36/5) â€“ x^{2}/4 = 1

On comparing this equation with the standard equation of hyperbola y^{2}/a^{2} â€“ x^{2}/b^{2} = 1,

We get a = 6/**âˆš**5 and b = 2,

It is known that, a^{2}Â + b^{2}Â = c^{2}

So,

c^{2}Â = 36/5Â + 4

c^{2} = 56/5

c = **âˆš**(56/5)

= 2**âˆš**14/**âˆš**5

Then,

The coordinates of the foci are (0, 2**âˆš**14/**âˆš**5) and (0, â€“ 2**âˆš**14/**âˆš**5).

The coordinates of the vertices are (0, 6/**âˆš**5) and (0, -6/**âˆš**5).

Eccentricity, e = c/a = (2**âˆš**14/**âˆš**5) / (6/**âˆš**5) = **âˆš**14/3

Length of latus rectum = 2b^{2}/a = (2 Ã— 2^{2})/6/âˆš5 = (2Ã—4)/6/**âˆš**5 = 4**âˆš**5/3

**6. 49y ^{2}Â â€“ 16x^{2}Â = 784.**

**Solution:**

Given:

The equation is 49y^{2}Â â€“ 16x^{2}Â = 784.

Let us divide the whole equation by 784, we get

49y^{2}/784 â€“ 16x^{2}/784 = 784/784

y^{2}/16 â€“ x^{2}/49 = 1

^{2}/a^{2} â€“ x^{2}/b^{2} = 1,

We get a = 4 and b = 7,

It is know that, a^{2}Â + b^{2}Â = c^{2}

So,

c^{2}Â = 16Â + 49

c^{2} = 65

c = **âˆš**65

Then,

The coordinates of the foci are (0, **âˆš**65) and (0, â€“**âˆš**65).

The coordinates of the vertices are (0, 4) and (0, -4).

Eccentricity, e = c/a = **âˆš**65/4

Length of latus rectum = 2b^{2}/a = (2 Ã— 7^{2})/4 = (2Ã—49)/4 = 49/2

**In each Exercises 7 to 15, find the equations of the hyperbola satisfying the given conditions**

**7. Vertices (Â±2, 0), foci (Â±3, 0)**

**Solution:**

Given:

Vertices (Â±2, 0) and foci (Â±3, 0)

Here, the vertices are on the x-axis.

So, the equation of the hyperbola is of the form x^{2}/a^{2} â€“ y^{2}/b^{2} = 1

Since, the vertices are (Â±2, 0), so, a = 2

Since, the foci are (Â±3, 0), so, c = 3

It is know that, a^{2}Â + b^{2}Â = c^{2}

So, 2^{2}Â + b^{2}Â = 3^{2}

b^{2}Â = 9 â€“ 4 = 5

âˆ´ The equation of the hyperbola is x^{2}/4 â€“ y^{2}/5 = 1

**8. Vertices (0,Â Â±Â 5), foci (0,Â Â±Â 8)**

**Solution:**

Given:

Vertices (0,Â Â±Â 5) and foci (0,Â Â±Â 8)

Here, the vertices are on the y-axis.

So, the equation of the hyperbola is of the form y^{2}/a^{2} â€“ x^{2}/b^{2} = 1

Since, the vertices areÂ (0,Â Â±5), so,Â a = 5

Since, the foci are (0,Â Â±8), so, c = 8

It is know that, a^{2}Â + b^{2}Â = c^{2}

So, 5^{2}Â + b^{2}Â = 8^{2}

b^{2}Â = 64 â€“ 25 = 39

âˆ´ The equation of the hyperbola is y^{2}/25 â€“ x^{2}/39 = 1

**9. Vertices (0,Â Â±Â 3), foci (0,Â Â±Â 5)**

**Solution:**

Given:

Vertices (0,Â Â±Â 3) and foci (0,Â Â±Â 5)

Here, the vertices are on the y-axis.

So, the equation of the hyperbola is of the formÂ y^{2}/a^{2} â€“ x^{2}/b^{2} = 1

Since, the vertices areÂ (0,Â Â±3), so,Â a = 3

Since, the foci are (0,Â Â±5), so, c = 5

It is known that, a^{2}Â + b^{2}Â = c^{2}

So, 3^{2}Â + b^{2}Â = 5^{2}

b^{2}Â = 25 â€“ 9 = 16

âˆ´ The equation of the hyperbola is y^{2}/9 â€“ x^{2}/16 = 1

**10. Foci (Â±5, 0), the transverse axis is of length 8.**

**Solution:**

Given:

Foci (Â±5, 0) and the transverse axis is of length 8.

Here, the foci are on x-axis.

The equation of the hyperbola is of the form x^{2}/a^{2} â€“ y^{2}/b^{2} = 1

Since, the foci are (Â±5, 0), so, c = 5

Since, the length of the transverse axis is 8,

2a = 8

a = 8/2

= 4

It is known that, a^{2}Â + b^{2}Â = c^{2}

4^{2}Â + b^{2}Â = 5^{2}

b^{2}Â = 25 â€“ 16

= 9

âˆ´ The equation of the hyperbola is x^{2}/16 â€“ y^{2}/9 = 1

**11. Foci (0,Â Â±13), the conjugate axis is of length 24.**

**Solution:**

Given:

Foci (0,Â Â±13) and the conjugate axis is of length 24.

Here, the foci are on y-axis.

The equation of the hyperbola is of the formÂ y^{2}/a^{2} â€“ x^{2}/b^{2} = 1

Since, the foci are (0,Â Â±13), so, c = 13

Since, the length of the conjugate axis is 24,

2b = 24

b = 24/2

= 12

It is known that, a^{2}Â + b^{2}Â = c^{2}

a^{2}Â + 12^{2}Â = 13^{2}

a^{2}Â = 169 â€“ 144

= 25

âˆ´ The equation of the hyperbola is y^{2}/25 â€“ x^{2}/144 = 1

**12. Foci (Â±Â 3âˆš5, 0), the latus rectum is of length 8.**

**Solution:**

Given:

Foci (Â±Â 3âˆš5, 0) and the latus rectum is of length 8.

Here, the foci are on x-axis.

The equation of the hyperbola is of the formÂ x^{2}/a^{2} â€“ y^{2}/b^{2} = 1

Since, the foci are (Â±Â 3âˆš5, 0), so, c =Â Â±Â 3âˆš5

Length of latus rectum is 8

2b^{2}/a = 8

2b^{2}Â = 8a

b^{2} = 8a/2

= 4a

It is known that, a^{2}Â + b^{2}Â = c^{2}

a^{2}Â + 4a = 45

a^{2}Â + 4a â€“ 45 = 0

a^{2}Â + 9a â€“ 5a â€“ 45 = 0

(a + 9) (a -5) = 0

a = -9 or 5

Since, a is non â€“ negative, a = 5

So, b^{2}Â = 4a

= 4 Ã— 5

= 20

âˆ´ The equation of the hyperbola is x^{2}/25 â€“ y^{2}/20 = 1

**13. Foci (Â±Â 4, 0), the latus rectum is of length 12**

**Solution:**

Given:

Foci (Â±Â 4, 0) and the latus rectum is of length 12

Here, the foci are on x-axis.

The equation of the hyperbola is of the formÂ x^{2}/a^{2} â€“ y^{2}/b^{2} = 1

Since, the foci areÂ (Â±Â 4, 0),Â so, c = 4

Length of latus rectum is 12

2b^{2}/a = 12

2b^{2}Â = 12a

b^{2} = 12a/2

= 6a

It is known that, a^{2}Â + b^{2}Â = c^{2}

a^{2}Â + 6a = 16

a^{2}Â + 6a â€“ 16 = 0

a^{2}Â + 8a â€“ 2a â€“ 16 = 0

(a + 8) (a â€“ 2) = 0

a = -8 or 2

Since, a is non â€“ negative, a = 2

So, b^{2}Â = 6a

= 6 Ã— 2

= 12

âˆ´ The equation of the hyperbola isÂ x^{2}/4 â€“ y^{2}/12 = 1

**14. Vertices (Â±7, 0), e = 4/3**

**Solution:**

Given:

Vertices (Â±7, 0) and e = 4/3

Here, the vertices are on the x- axis

The equation of the hyperbola is of the form x^{2}/a^{2} â€“ y^{2}/b^{2} = 1

Since, theÂ vertices are (Â±Â 7, 0), so, a = 7

It is given that e =Â 4/3

c/a = 4/3

3c = 4a

Substitute the value of a, we get

3c = 4(7)

c = 28/3

It is known that, a^{2}Â + b^{2}Â = c^{2}

7^{2}Â + b^{2}Â = (28/3)^{2}

b^{2} = 784/9 â€“ 49

= (784 â€“ 441)/9

= 343/9

âˆ´ The equation of the hyperbola is x^{2}/49 â€“ 9y^{2}/343 = 1

**15. Foci (0,Â Â±âˆš10), passing through (2, 3)**

**Solution:**

Given:

Foci (0,Â Â±âˆš10) and passing through (2, 3)

Here, the foci are on y-axis.

The equation of the hyperbola is of the form y^{2}/a^{2} â€“ x^{2}/b^{2} = 1

Since, the foci areÂ (Â±âˆš10, 0),Â so, c =Â âˆš10

It is known that, a^{2}Â + b^{2}Â = c^{2}

b^{2}Â = 10 â€“ a^{2}Â â€¦â€¦â€¦â€¦.. (1)

It is given that the hyperbola passes through point (2, 3)

So, 9/a^{2} â€“ 4/b^{2} = 1 â€¦ (2)

From equations (1) and (2), we get,

9/a^{2} â€“ 4/(10-a^{2}) = 1

9(10 â€“ a^{2}) â€“ 4a^{2}Â = a^{2}(10 â€“a^{2})

90 â€“ 9a^{2}Â â€“ 4a^{2}Â = 10a^{2}Â â€“ a^{4}

a^{4}Â â€“ 23a^{2}Â + 90 = 0

a^{4}Â â€“ 18a^{2}Â â€“ 5a^{2 }+ 90 = 0

a^{2}(a^{2}Â -18) -5(a^{2}Â -18) = 0

(a^{2}Â â€“ 18) (a^{2}Â -5) = 0

a^{2}Â = 18 or 5

In hyperbola, c > a i.e., c^{2 }> a^{2}

So, a^{2}Â = 5

b^{2}Â = 10 â€“ a^{2}

= 10 â€“ 5

= 5

âˆ´ The equation of the hyperbola is y^{2}/5 â€“ x^{2}/5 = 1

Miscellaneous EXERCISE PAGE NO: 264

**1. If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.**

**Solution:**

We know that the origin of the coordinate plane is taken at the vertex of the parabolic reflector, where the axis of the reflector is along the positive x â€“ axis.

Diagrammatic representation is as follows:

We know that the equation of the parabola is of the form y^{2}Â = 4ax (as it is opening to the right)

Since, the parabola passes through point A(10, 5),

y^{2}Â = 4ax

10^{2}Â = 4a(5)

100 = 20a

a = 100/20

= 5

The focus of the parabola is (a, 0) = (5, 0), which is the mid â€“ point of the diameter.

Hence, the focus of the reflector is at the mid-point of the diameter.

**2. An arch is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?**

**Solution:**

We know that the origin of the coordinate plane is taken at the vertex of the arch, where its vertical axis is along the positive y â€“axis.

Diagrammatic representation is as follows:

The equation of the parabola is of the form x^{2}Â = 4ay (as it is opening upwards).

It is given that at base arch is 10m high and 5m wide.

So, y = 10 and x = 5/2 from the above figure.

It is clear that the parabola passes through point (5/2, 10)

So, x^{2}Â = 4ay

(5/2)^{2} = 4a(10)

4a = 25/(4Ã—10)

a = 5/32

we know the arch is in the form of a parabola whose equation is x^{2} = 5/8y

We need to find width, when height = 2m.

To find x, when y = 2.

When, y = 2,

x^{2} = 5/8 (2)

= 5/4

x = âˆš(5/4)

= âˆš5/2

AB = 2 Ã— âˆš5/2m

= âˆš5m

= 2.23m (approx.)

Hence, when the arch is 2m from the vertex of the parabola, its width is approximately 2.23m.

**3. The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 100 m long is supported by vertical wires attached to the cable, the longest wire being 30 m and the shortest being 6 m.
Find the length of a supporting wire attached to the roadway 18 m from the middle.**

**Solution:**

We know that the vertex is at the lowest point of the cable. The origin of the coordinate plane is taken as the vertex of the parabola, while its vertical axis is taken along the positive y â€“axis.

Diagrammatic representation is as follows:

Here, AB and OC are the longest and the shortest wires, respectively, attached to the cable.

DF is the supporting wire attached to the roadways, 18m from the middle.

So, AB = 30m, OC = 6m, and BC = 50m.

The equation of the parabola is of the from x^{2} = 4ay (as it is opening upwards).

The coordinates of point A are (50, 30 -6) = (50, 24)

Since, A(50, 24) is a point on the parabola.

y^{2}Â = 4ax

(50)^{2}Â = 4a(24)

a = (50Ã—50)/(4Ã—24)

= 625/24

Equation of the parabola,Â x^{2} = 4ay = 4Ã—(625/24)y or 6x^{2} = 625y

The x â€“ coordinate of point D is 18.

Hence, at x = 18,

6(18)^{2}Â = 625y

y = (6Ã—18Ã—18)/625

= 3.11(approx.)

Thus, DE = 3.11 m

DF = DE +EF = 3.11m +6m = 9.11m

Hence, the length of the supporting wire attached to the roadway 18m from the middle is approximately 9.11m.

**4. An arch is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.**

**Solution:**

Since, the height and width of the arc from the centre is 2m and 8m respectively, it is clear that the length of the major axis is 8m, while the length of the semi- minor axis is 2m.

The origin of the coordinate plane is taken as the centre of the ellipse, while the major axis is taken along the x-axis.

Hence, Diagrammatic representation of semi- ellipse is as follows:

The equation of the semi â€“ ellipse will be of the from x^{2}/16 + y^{2}/4 = 1, y â‰¥ 0 â€¦ (1

Let A be a point on the major axis such that AB = 1.5m.

Now draw AC âŠ¥ OB.

OA = (4 â€“ 1.5)m = 2.5m

The x â€“ coordinate of point C is 2.5

On substituting the value of x with 2.5 in equation (1), we get,

(2.5)^{2}/16 + y^{2}/4 = 1

6.25/16 + y^{2}/4 = 1

y^{2} = 4 (1 â€“ 6.25/16)

= 4 (9.75/16)

= 2.4375

y = 1.56 (approx.)

So, AC = 1.56m

Hence, the height of the arch at a point 1.5m from one end is approximately 1.56m.

**5. A rod of length 12 cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point P on the rod, which is 3 cm from the end in contact with the x-axis.**

**Solution:**

Let AB be the rod making an angleÂ ÆŸÂ with OX and P(x,y) be the point on it such that

AP = 3cm.

Diagrammatic representation is as follows:

Then, PB = AB â€“ AP = (12 â€“ 3) cm = 9cm [AB = 12cm]

From P, draw PQÂ âŠ¥ OY and PR âŠ¥ OX.

In Î”PBQ,Â cos Î¸ = PQ/PB = x/9

Sin Î¸ = PR/PA = y/3

we know that, sin^{2} Î¸Â +cos^{2} Î¸Â = 1,

So,

(y/3)^{2} + (x/9)^{2} = 1 or

x^{2}/81 + y^{2}/9 = 1

Hence, the equation of the locus of point P on the rod is x^{2}/81 + y^{2}/9 = 1

**6. Find the area of the triangle formed by the lines joining the vertex of the parabola x ^{2}Â = 12y to the ends of its latus rectum.**

**Solution:**

The given parabola is x^{2}Â = 12y.

On comparing this equation with x^{2} = 4ay, we get,

4a = 12

a = 12/4

= 3

The coordinates of foci are S(0,a) = S(0,3).

Now let AB be the latus rectum of the given parabola.

The given parabola can be roughly drawn as

At y = 3, x^{2}Â = 12(3)

x^{2}Â = 36

x =Â **Â±**6

So, the coordinates of A are (-6, 3), while the coordinates of B are (6, 3)

Then, the vertices of Î”OAB are O(0,0), A (-6,3) and B(6,3).

By using the formula,

Area of Î”OAB = Â½ [0(3-3) + (-6)(3-0) + 6(0-3)] unit^{2}

= Â½ [(-6) (3) + 6 (-3)] unit^{2}

= Â½ [-18-18] unit^{2}

= Â½ [-36] unit^{2}

= 18 unit^{2}

âˆ´ Area of Î”OAB is 18 unit^{2}

**7. A man running a racecourse notes that the sum of the distances from the two flag posts from him is always 10 m and the distance between the flag posts is 8 m.
Find the equation of the posts traced by the man.**

**Solution:**

Let A and B be the positions of the two flag posts and P(x, y) be the position of the man.

So, PA + PB = 10.

We know that if a point moves in plane in such a way that the sum of its distance from two fixed point is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.

Then, the path described by the man is an ellipse where the length of the major axis is 10m, while points A and B are the foci.

Now let us take the origin of the coordinate plane as the centre of the ellipse, and taking the major axis along the x- axis,

The diagrammatic representation of the ellipse is as follows:

The equation of the ellipse is in the form of x^{2}/a^{2} + y^{2}/b^{2} = 1, where â€˜aâ€™ is the semi-major axis.

So, 2a = 10

a = 10/2

= 5

Distance between the foci, 2c = 8

c = 8/2

= 4

By using the relation, c =** âˆš**(a^{2} â€“ b^{2}), we get,

4 = **âˆš**(25 â€“ b^{2})

16 = 25 â€“ b^{2}

b^{2}Â = 25 -1

= 9

b = 3

Hence, equation of the path traced by the man is x^{2}/25 + y^{2}/9 = 1

**8. An equilateral triangle is inscribed in the parabola y ^{2}Â = 4ax, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.**

**Solution:**

Let us consider OAB be the equilateral triangle inscribed in parabola y^{2}Â = 4ax.

Let AB intersect the x â€“ axis at point C.

Diagrammatic representation of the ellipse is as follows:

Now let OC = k

From the equation of the given parabola, we have,

So, y^{2}Â = 4ak

y =** Â±**2**âˆš**ak

The coordinates of points A and B are (k, 2**âˆš**ak), and (k, -2**âˆš**ak)

AB = CA + CB

=Â 2**âˆš**ak + 2**âˆš**ak

= 4**âˆš**ak

Since, OAB is an equilateral triangle, OA^{2}Â = AB^{2}.

Then,

k^{2} + (2**âˆš**ak)^{2} = (4**âˆš**ak)^{2}

k^{2}Â + 4ak = 16ak

k^{2}Â = 12ak

k = 12a

Thus, AB =Â 4**âˆš**ak = 4**âˆš**(aÃ—12a)

= 4**âˆš**12a^{2}

= 4**âˆš**(4aÃ—3a)

= 4(2)**âˆš**3a

= 8**âˆš**3a

Hence, the side of the equilateral triangle inscribed in parabola y^{2}Â = 4ax isÂ 8**âˆš**3a.

## NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections

Chapter 11 Conic Sections of NCERT Solutions for Class 11 covers the topics listed below:

11.1 Introduction

11.2 Sections of a Cone

11.2.1 Circle, ellipse, parabola and hyperbola

11.2.2 Degenerated conic sections

11.3 Circle

11.4 Parabola

11.4.1 Standard equations of parabola

11.4.2 Latus rectum

11. 5 Ellipse

11.5.1 Relationship between semi-major axis, semi-minor axis and the distance of the focus from the centre of the ellipse

11.5.2 Special cases of an ellipse

11.5.3 Eccentricity

11.5.4 Standard equations of an ellipse

11.5.5 Latus rectum

11.6 Hyperbola

11.6.1 Eccentricity

11.6.2 Standard equation of Hyperbola

11.6.3 Latus rectum

Exercise 11.1 Solutions 15 Questions

Exercise 11.2 Solutions 12 Questions

Exercise 11.3 Solutions 20 Questions

Exercise 11.4 Solutions 15 Questions

Miscellaneous Exercise On Chapter 11 Solutions 8 Questions

## NCERT Solutions for Class 11 Maths Chapter 11- Conic Sections

The chapter Conic Sections belongs to the unit Coordinate Geometry, that adds up to 10 marks of the total 80 marks. The number of exercises is four and a miscellaneous exercise, adding to 5 exercises in total. Provided below are the concepts covered in this Chapter of NCERT Solutions for Class 11 Maths.

- A circle is the set of all points in a plane that are equidistant from a fixed point in the plane.
- The equation of a circle with centre (h, k) and the radius r is (x â€“ h)
^{2}+ (y â€“ k)^{2}= r^{2}. - A parabola is the set of all points in a plane that are equidistant from a fixed line and a fixed point in the plane.
- The equation of the parabola with focus at (a, 0) a > 0 and directrix x = â€“ a is y
^{2}= 4ax. - Latus rectum of a parabola is a line segment perpendicular to the axis of the parabola, through the focus and whose endpoints lie on the parabola.
- Length of the latus rectum of the parabola y
^{2}= 4ax is 4a. - An ellipse is the set of all points in a plane, the sum of whose distances from two fixed points in the plane is constant.
- The equation of an ellipse with foci on the x-axis.
- Latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose endpoints lie on the ellipse.
- Length of the latus rectum of the ellipse.
- The eccentricity of an ellipse is the ratio between the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse.
- A hyperbola is the set of all points in a plane, the difference of whose distances from two fixed points in the plane is constant.
- The equation of a hyperbola with foci on the x-axis.
- Latus rectum of a hyperbola is a line segment perpendicular to the transverse axis through any of the foci and whose endpoints lie on the hyperbola.
- Length of the latus rectum of the hyperbola.
- The eccentricity of a hyperbola is the ratio of the distances from the centre of the hyperbola to one of the foci and to one of the vertices of the hyperbola.

Studying the Conic Sections of Class 11 enables the students to get a strong knowledge of the concepts of Sections of a cone: circles, ellipse, parabola, hyperbola, a point, a straight line and a pair of intersecting lines as a degenerate case of a conic section. The students would also be able to understand the standard equations and simple properties of parabola, ellipse, hyperbola and circle.