**Exercise 11.1**** **

**Q.1: A circle is given with centre (0, 3) and radius 3. Obtain the equation of the given circle.**

** **

**Sol:**

**The General equation of a circle with centre (h, k) and radius r is:**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

**Given:**

**(h, k) = centre of a circle = (0, 3) and radius r = 3**

**The equation of a circle is:**

(a – 0)^{2} + (b – 3)^{2} = 3^{2}

a^{2 }+ b^{2} – 6b + 9 = 9

a^{2 }+ b^{2} – 6b + 9 – 9 = 0

**a ^{2 }+ b^{2} – 6b = 0**

** **

**Q.2: A circle is given with centre (-1, 2) and radius 4. Find the equation of the circle.**

** **

**Sol:**

**The General equation of a circle with centre (h, k) and radius r is:**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

**Given:**

**Centre of a circle = (h, k) = (-1, 2) and radius (r) = 4**

**The equation of a circle is:**

(a – (- 1))^{2} + (b – 2)^{2} = 4^{2}

(a + 1)^{2} + (b – 2)^{2} = 4^{2}

a^{2 }+ 2a + 1 + b^{2} – 4b + 4 = 16

a^{2 }+ 2a + b^{2} – 4b + 5 = 16

a^{2 }+ 2a + b^{2} – 4b + 5 – 16 = 0

**a ^{2 }+ 2a + b^{2} – 4b – 11 = 0**

** **

**Q.3: A circle is given with centre (\(\frac{1}{3}\), ****\(\frac{1}{2}\)) and radius \(\frac{1}{14}\). Find the equation of the circle.**

** **

**Sol:**

**The General equation of a circle with centre (h, k) and radius r is:**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

**Given:**

Centre of a circle = (h, k) = (\(\frac{1}{3}\), \(\frac{1}{2}\)) and the radius (r) = 4

**The equation of a circle is:**

(a – \(\frac{1}{3}\))^{2} + (b – \(\frac{1}{2}\))^{2} = \(\frac{1}{14}\)^{2}

(a – \(\frac{1}{3}\))^{2} + (b – \(\frac{1}{2}\))^{2} = \(\frac{1}{14}\)^{2}

**Q.4: A circle is given with centre (2, 2) and radius \(\sqrt{5}\). Obtain the equation of the given circle.**

** **

**Sol:**

**The General equation of a circle with centre (h, k) and radius r is:**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

Given:

Centre of a circle = (h, k) = (2, 2) and radius (r) = \(\sqrt{5}\)

**The equation of a circle is:**

(a – 2)^{2} + (b – 2)^{2} = \(\sqrt{5}^{2}\)

(a – 2)^{2} + (b – 2)^{2} = \(\sqrt{5}^{2}\)

a^{2 }– 4a + 4 + b^{2} – 4b + 4 = 5

a^{2 }– 4a + b^{2} – 4b + 8 = 5

a^{2 }– 4a + b^{2} – 4b + 8 – 5 = 0

a^{2 }– 4a + b^{2} – 4b + 3 = 0

**a ^{2 }+ b^{2} – 4a – 4b + 3 = 0**

** **

**Q.5: A circle is given with centre (- x, – y) and radius \(\sqrt{x^{2} – y^{2}}\). Find the equation of the circle.**

** **

**Sol:**

**The General equation of a circle with centre (h, k) and radius r is:**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

Given:

Centre of a circle = (h, k) = (- x, – y) and radius ‘r’ = \(\sqrt{x^{2} – y^{2}}\)

**The equation of a circle is:**

(a – (- x)^{2} + (b – (- y)^{2} = \(\sqrt{x^{2} – y^{2}}^{2}\)

(a + x)^{2} + (b + y)^{2} = \(\sqrt{x^{2} – y^{2}}^{2}\)

(a + x)^{2} + (b + y)^{2} = x^{2} – y^{2}

a^{2 }+ 2ax + x^{2} + b^{2} + 2by + y^{2} = x^{2} – y^{2}

**a ^{2 }+ 2ax + b^{2} + 2by + y^{2} = 0**

** **

**Q.6: The equation of a given circle is given as (a + 5) ^{2} + (b – 3)^{2} = 36. Find the centre and radius of the circle.**

** **

**Sol:**

Given:

The equation of a given circle (a + 5)^{2} + (b – 3)^{2} = 36

(a + 5)^{2} + (b – 3)^{2} = 36

{a – (- 5)}^{2} + (b – 3) = 6^{2}

Which is of the form (a – h) ^{2 }+ (b – k) ^{2} = r^{2}, where h = – 5, k = 3 and r = 6.

**Hence, the centre of the circle is (-5, 3), and the radius of circle is 6.**

**Q.7: The equation of a given circle is given as a ^{2 }+ b^{2} – 4a – 4b + 3 = 0**

**Find the radius and centre of the circle.**

** **

**Sol:**

Given:

The equation of a given circle a^{2 }+ b^{2} – 4a – 4b + 3 = 0

a^{2 }+ b^{2} – 4a – 4b + 3 = 0

a^{2 }– 4a + b^{2} – 4b + 3 = 0

{a^{2 }– 2.a.2 + 2^{2}} + {b^{2 }– 2.b.2 + 2^{2}} – 2^{2} – 2^{2} + 3 = 0

{a^{2 }– 2.a.2 + 2^{2}} + {b^{2 }– 2.b.2 + 2^{2}} – 8 + 3 = 0

{a^{2 }– 2.a.2 + 2^{2}} + {b^{2 }– 2.b.2 + 2^{2}} = 5

(a – 2)^{2} + (b – 2)^{2} = \(\sqrt{5}^{2}\)

Which is of the form (a – h) ^{2 }+ (b – k) ^{2} = r^{2}, where h = 2, k = 2 and r = \(\sqrt{5}\).

**Thus, the centre of the circle is (2, 2), and the radius of circle is \(\sqrt{5}\).**

** **

**Q.8: The equation of a given circle is given as a ^{2 }+ b^{2} – 6a + 8b – 14 = 0**

**Find the centre and radius of the circle.**

** **

**Sol:**

**Given:**

** The equation of a given circle a ^{2 }+ b^{2} – 6a + 8b – 14 = 0**

a^{2 }– 6a + b^{2} + 8b – 14 = 0

{a^{2 }– 2.a.3 + 3^{2}} + {b^{2} + 2.b.4 + 4^{2}} – 3^{2 }– 4^{2 }– 14 = 0

(a – 3) ^{2} + (b + 4) ^{2} – 9 – 16 – 14 = 0

(a – 3) ^{2} + (b + 4) ^{2} – 39 = 0

(a – 3) ^{2} + (b + 4) ^{2} = 39

(a – 3) ^{2} + (b – (- 4)) ^{2} = 39

Which is of the form (a – h) ^{2 }+ (b – k) ^{2} = r^{2}, where h = 3, k = 4 and r = \(\sqrt{39}\).

**Thus, the centre of the circle is (3, 4), and the radius of circle is \(\sqrt{39}\).**

** **

**Q.9: The equation of a given circle is given as 2a ^{2 }+ 2b^{2} – 8a = 0**

**Find the centre and radius of the circle.**

** **

**Sol:**

Given:

The equation of a given circle 2a^{2 }+ 2b^{2} – 8a = 0

2a^{2 }+ 2b^{2} – 8a = 0

2a^{2 }– 8a + 2b^{2} = 0

2[a^{2} – 4a + b^{2}] = 0

{a^{2} – 2.a.2 + 2^{2}} – 2^{2} + (b – 0)^{2} = 0

[a^{2} – 2.a.2 + 2^{2}] + [b – 0]^{2} – 2^{2} = 0

(a – 2)^{2} + (b – 0)^{2} = 2^{2} = 4

Which is of the form (a – h) ^{2 }+ (b – k) ^{2} = r^{2}, where h = 2, k = 0 and r = \(\sqrt{4}\) = 2.

**Thus, the centre of the circle is (3, 4), and the radius of circle is 2.**

** **

**Q.10: The circle passing through the points (6, 2) and (4, 3) and whose centre lies on the line 2a + y = 8. Find the equation of the circle.**

** **

**Sol:**

**The equation of a required circle with centre (h, k) and radius r is given as**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

As the circle passes through (6, 2) and (4, 3)

(6 – h)^{2} + (2 – k)^{2} = r^{2} **. . . . . . . . . . . . . . . . (1)**

(4 – h)^{2} + (3 – k)^{2} = r^{2} **. . . . . . . . . . . . . . . . (2)**

The centre (h, k) of the circle lies on the line 2a + y = 8,

2h + k = 8 **. . . . . . . . . . . . . . . . . . . . . (3)**

**Now, On Comparing equations 1 and 2, we will get:**

(6 – h)^{2} + (2 – k)^{2} = (4 – h)^{2} + (3 – k)^{2}

36 – 12h + h^{2} + 4 – 4k + k^{2} = 16 – 8h + h^{2} + 9 – 6k + k^{2}

36 – 12h + 4 – 4k = 16 – 8h + 9 – 6k

– 12h – 4k + 8h + 6k = 16 + 9 – 36 – 4

4h – 2k = 15 **. . . . . . . . . . . . . . . . . . (4)**

**Now, on Solving Equations (3) and (4), we will get:**

h = \(\frac{31}{8}\)

k = \(\frac{1}{4}\)

**Substituting the values of h and k in equation (1), we will get:**

(6 – \(\frac{31}{8}\))^{2} + (2 – \(\frac{1}{4}\))^{2} = r^{2}

\(\\\Rightarrow\) \(\frac{289}{64} + \frac{49}{16} = r^{2}\)

\(\\\Rightarrow\) \(\frac{289 + 196}{64} = r^{2}\)

\(\\\Rightarrow\) \(\frac{485}{64} = r^{2}\)

**\(\Rightarrow\) r = 7.57**

**The required equation is:**

(a – \(\frac{31}{8}\))^{2} + (b – \(\frac{1}{4}\))^{2} = r^{2}

(a – \(\frac{31}{8}\))^{2} + (b – \(\frac{1}{4}\))^{2} = 7.57^{2}

\(\\\Rightarrow\) \(a^{2} – \frac{31}{4}a + (\frac{31}{8})^{2} + b^{2} – \frac{b}{2} + \frac{1}{16} = 57.30\\\)

\(\\\Rightarrow\) \(a^{2} – \frac{31}{4}a + \frac{961}{64} + b^{2} – \frac{b}{2} + \frac{1}{16} =57.30\)

**Therefore, the required equation is: 64a ^{2}– 496a + 64b^{2} -32b = 2702.51**

**Q.11: The circle passing through the points (3, 2) and (-2, 2) and whose centre lies on the line a – 3y – 13 = 0. Find the equation of the circle.**

** **

**Sol:**

**The General equation of a circle with centre (h, k) and radius r is:**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

As the circle passes through (3, 2) and (-2, 1)

**(3 – h) ^{2} + (2 – k)^{2} = r^{2} . . . . . . . . . . . . . . . . (1)**

**(-2 – h) ^{2} + (2 – k)^{2} = r^{2} . . . . . . . . . . . . . . . (2)**

The centre (h, k) of the circle lies on the line a – 3y – 13 = 0,

**h – 3k = 13 . . . . . . . . . . . . . . . (3)**

**Now, on comparing equations 1 and 2, we will get:**

(3 – h)^{2} + (2 – k)^{2 }= (-2 – h)^{2} + (2 – k)^{2}

9 – 6h + h^{2} = 4 + 4h + h^{2}

9 – 6h = 4 + 4h

**h = 0.5 . . . . . . . . . . . . . . . . . . (4)**

**Now, on solving equation (3), we will get:**

k = – \(\frac{25}{6}\) = – 4.166

**Now, on substituting the values of h and k in equation (1), we will get:**

(3 – 0.5)^{2} + (2 – (- \(\frac{25}{6}\)))^{2} = r^{2}

(2.5) ^{2} + (2 + \(\frac{25}{6}\))^{2} = r^{2}

(2.5) ^{2} + (2 + 4.16)^{2} = r^{2}

(2.5) ^{2} + (6.16)^{2} = r^{2}

r^{2 }= 44.1956

r = 6.65

**The required equation is:**

(a – h)^{2} + (b – k)^{2} = r^{2}

(a – 0.5)^{2} + (b – (- 4.16))^{2} = r^{2}

(a – 0.5)^{2} + (b + 4.16)^{2} = r^{2}

a^{2} – a + 6.25 + b^{2} + 8.32b + 17.30 = 44.19

a^{2} – a + b^{2} + 8.32b = 44.19 – 6.25 – 17.30

**a ^{2} + b^{2} – a + 8.32b – 20.64 = 0**

** **

**Q.12: The radius of the circle is 6 whose centre lies on x – axis and passes through the point (3, 2). Find the equation.**

** **

**Sol:**

Given:

The radius of the circle is 6 and passes through the point (3, 2)

**The General equation of a circle with centre (h, k) and radius r is:**

**(a – h) ^{2} + (b – k)^{2} = r^{2}**

As, the centre lies on x – axis, k = 0, the equation becomes:

(a – h)^{ 2} + b^{ 2} = 36

(3 – h)^{ 2} + 2^{2} = 36

(3 – h)^{ 2} = 36 – 4

3 – h = \(\sqrt{32}\)

h = \(3 + 4\sqrt{2}\) or h = \(3 – 4\sqrt{2}\)

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

** **

**Sol:**

**Suppose, the equation of the required circle be (x – h) ^{2} + (y – k) ^{2} = r ^{2}**

The centre of the circle passes through (0, 0):

(0 – h) ^{2} + (0 – k) ^{2 }= r ^{2}

h ^{2} + k ^{2} = r ^{2}

Therefore, the equation of the circle is:

(x – h) ^{2} + (y – k) ^{2} = h ^{2} + k ^{2} .

Given, the circle makes intercepts a and b on the coordinate axes.

The circle passes through points (a, 0) and (0, b).

Therefore, (a – h) ^{2} + (0 – k) ^{2} = h ^{2} + k ^{2} ** . . . . . . . . . . . . . . . (1)**

(0 – h) ^{2} + (b – k) ^{2} = h ^{2} + k ^{2} **. . . . . . . . . . . . . . . . . . (2)**

Now, from equation (1), we obtain:

a ^{2} – 2ah + h ^{2} + k ^{2} = h ^{2} + k ^{2}

a^{ 2} – 2ah = 0

a (a – 2h) = 0

a = 0 or (a – 2h) = 0

We know, a ≠ 0; hence, (a – 2h) = 0 ⇒ h =a/2.

Now, from equation (2), we will get:

h ^{2} + b ^{2} – 2bk + k ^{2} = h ^{2} + k ^{2}

b ^{2} – 2bk = 0

b (b – 2k) = 0

b = 0 or(b – 2k) = 0

However, b ≠ 0;

Therefore, (b – 2k) = 0 ⇒ **k =b/2.**

**Thus, the equation of the required circle is:**

**Q.14: ****The circle with centre (3, 3) and passes through the point (6, 5), find the equation of a circle?**

** **

**Sol:**

**Given:**

**The centre of the circle (h, k) = (3, 3). **

As the circle passes through point (6, 5), the radius (r) of the circle is the distance between the points (2, 2) and (6, 5).

\(r = \sqrt{(3 – 6)^{2} + (3 – 5)^{2}} + \sqrt{(- 3)^{2} + (- 2)^{2}} = \sqrt{9 + 4} = \sqrt{13}\\\)**Therefore, the equation of the circle is**:

**Q.****15: Check whether the point (–2.0, 3.0) lies inside, outside or on the circle x ^{2} + y ^{2} = 25? **

** **

**Sol:**

**Given:**

**The equation of the given circle is x ^{2} + y ^{2} = 25**

x ^{2} + y ^{2} = 25

(x – 0)^{2} + (y – 0)^{2} = 5^{2} , which is of the form (a – h)^{ 2} + (b – k)^{ 2} = r^{2}, where h = 0, k = 0, and r = 5.

Centre = (0, 0) and radius = 5

Distance between point (– 2.0, 3.0) and centre (0, 0)

\(\sqrt{(- 2 – 0)^{2} + (3 – 0)^{2}}\\\)=\(\sqrt{4 + 9} = \sqrt{13} = 3.60 (approx.) < 5\)

As, the distance between point (– 2.0, 3.0) and centre (0, 0) of the circle is less than the radius of the circle, **point (– 2.0, 3.0) lies inside the circle.**

** **

**Exercise 11.2**

**Q.1: For the equation y ^{2} = 16x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.**

** **

**Sol:**

**Given:**

** y ^{2} = 16x**

As the coefficient of x is positive, the given **parabola has the opening on the right side.**

By comparing this equation with y^{2} = 4ax, we get:

4a = 16 ⇒ **a = 4**

Therefore, **focus coordinates** = (a, 0) **= (4, 0)**

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

Directrix equation, x = – a = – 4

So, **x + 4 = 0**

**Therefore, Length of the latus rectum = 4a = 4 × 4 = 16**

**Q.2: For the equation x ^{2} = 8y, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.**

** **

**Sol:**

**Given:**

** x ^{2} = 8y**

As the coefficient of y is positive, the given** parabola has the opening upwards.**

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

4a = 8 ⇒ **a = 2**

Therefore, **focus coordinates** = (0, a) **= (0, 2)**

The given equation has x^{2}, the y – axis is the axis of the parabola.

Directrix equation, y = – a = – 2

So, **x + 2 = 0**

**Therefore, Length of the latus rectum = 4a = 4 × 2 = 8**

** **

**Q.3: For the equation y ^{2} = – 12x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.**

** **

**Sol:**

**Given:**

** y ^{2} = – 12x**

As the coefficient of x is negative, the given **parabola has the opening on the left side.**

By comparing this equation with y^{2} = – 4ax, we get

4a = 12 ⇒ **a = 3**

Therefore, **focus coordinates** = (- a, 0) **= (- 3, 0)**

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

** Directrix equation, x = a = 2**

So,** x + 4 = 0**

**Therefore, Length of the latus rectum = 4a = 4 × 3 = 12**

** **

**Q.4: For the equation x ^{2} = – 20y, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.**

** **

**Sol:**

**Given:**

** x ^{2} = – 20y**

As the coefficient of y is negative, the given **parabola has the opening downwards.**

By comparing this equation with x^{2} = – 4ay, we get:

– 4a = – 20 ⇒ **a = 5**

Therefore, **focus coordinates** = (0, a) = **(0, 5)**

The given equation has x^{2}, the y – axis is the axis of the parabola.

** Directrix equation, y = a = 5**

**Therefore, Length of the latus rectum = 4a = 4 × 5 = 20**

** **

**Q.5: For the equation y ^{2} = 24x, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.**

** **

**Sol:**

**Given:**

** y ^{2} = 24x**

As the coefficient of x is positive, the given **parabola has the opening on the right side.**

By comparing this equation with y^{2} = 4ax, we get:

4a = 24 ⇒ **a = 6**

Therefore, **focus coordinates**** =** (a, 0) = **(6, 0)**

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

**Directrix equation, x = – a = – 6**

**So, x + 6 = 0**

**Therefore, Length of the latus rectum = 4a = 4 × 6 = 24**

**Q.6: For the equation x ^{2} = – 7y, determine the focus coordinates, the directrix equation, the length of the latus rectum and the axis of the parabola.**

**Sol:**

**Given:**

** x ^{2} = – 7y**

As the coefficient of y is negative, the given **parabola has the opening downwards.**

By comparing this equation with x^{2} = – 4ay, we get:

– 4a = – 7 ⇒ a = \(\frac{7}{4}\)

Therefore, **focus coordinates** = (0, a) = (0, \(\frac{7}{4}\))

The given equation has x^{2}, **the y – axis is the axis of the parabola**.

** Directrix equation,** y = a = \(\frac{7}{4}\)

**Therefore, Length of the latus rectum = 4a = 4 × \(\frac{7}{4}\) = 7**

** **

**Q7. Obtain the equation of the parabola which satisfies the given conditions below: **

**(i) Focus coordinates**** = (8, 0) **

**(ii) Directrix (x) = – 8**

** **

**Sol:**

**Given:**

**(i) Focus coordinates**** = (8, 0)**

**(ii) Directrix (x) = – 8**

The axis of the given parabola is the x – axis as the focus is represented on the x – axis.

The required equation can be either y^{2} = 4ax and y^{2} = – 4ax

As we know, **d****irectrix (x) = – 8**

As the coefficient of x is negative, the given **parabola has the opening on the** **left side and the focus (8, 0)**

Therefore, the equation of the parabola = y^{2} = 4ax,

**Here, a = 8**

**Thus, the equation of the parabola = y ^{2} = 32 x.**

** **

**Q.8: Obtain the equation of the parabola which satisfies the given conditions below: **

**(i) Focus coordinates**** = (0, – 5)**

**(ii) Directrix (y) = 5**

** **

**Sol:**

**Given:**

**(i) Focus coordinates**** = (0, – 5)**

**(ii) Directrix (x) = 5**

The axis of the given parabola is the y – axis as the focus is represented on the y – axis.

The required equation can be either x^{2} = 4ay and x^{2} = – 4ay

As we know, **d****irectrix (x) = 5**

As the coefficient of x is negative, the given **parabola has the opening on the left side and the focus (0, -5)**

Therefore, the equation of the parabola = x^{2} = – 4ay,

**Here, a = 5**

**Thus, the equation of the parabola = x ^{2} = – 20y**

**Q.9: Obtain the equation of the parabola which satisfies the given conditions below: **

**(i) Vertex is at origin.**

**(ii) Focus coordinates**** = (4, 0)**

** **

**Sol:**

**Given:**

**(i) Vertex is at origin i.e., (0, 0)**

**(ii) Focus coordinates**** = (4, 0)**

As the vertex of a parabola lies at origin and the focus is represented on the positive side of the x – axis, and hence, the axis of the parabola is x – axis.

**The required equation is ****y ^{2} = 4ax **

As we know, focus is (4, 0)

Therefore, the equation of the parabola = y^{2} = 4ax

**Here, a = 4**

**Thus, the equation of the parabola = y ^{2} = 16 x **

** **

**Q.10: Obtain the equation of the parabola which satisfies the given conditions below: **

**(i) Vertex is at origin.**

**(ii) Focus coordinates**** = (- 3, 0)**

** **

**Sol:**

**Given:**

**(i) Vertex is at origin i.e., (0, 0)**

**(ii) Focus coordinates**** = (- 3, 0)**

As the vertex of a parabola lies at origin and the focus is represented on the negative side of the x – axis, and hence, **the axis of the parabola is x – axis.**

The required equation is **y ^{2} = – 4ax**

Therefore, the equation of the parabola = y^{2} = – 4ax

As we know, focus is (- 3, 0)

**Here, a = 3**

**Thus, the equation of the parabola = y ^{2} = – 12 x**

** **

**Q.11: Obtain the equation of the parabola which satisfies the given conditions below: **

**(i) Vertex is at origin.**

**(ii) Passing through coordinates**** = (3, 5)**

**The axis of the parabola is on x – axis.**

**Sol:**

**Given:**

**(i) Vertex is at origin i.e., (0, 0)**

**(ii) Passing through coordinates**** = (3, 5) **

As the vertex of a parabola lies at origin and the passing through coordinates = (3, 5).

The required equation is either y^{2} = 4ax or y^{2} = – 4ax

**The coordinates (3, 5) is in first quadrant**.

Therefore, the equation of the parabola = y^{2} = 4ax, whereas (3, 5) should satisfy the equation

So,

5^{2} = 4a (3)

25 = 12 a

a = \(\frac{25}{12}\)

**The parabola’s equation is: **

y^{2} = 4 (\(\frac{25}{12}\)) x

**3 y ^{2} = 25 x**

** **

**Q.12: Obtain the equation of the parabola which satisfies the given conditions below: **

**(i) Vertex is at origin.**

**(ii) Passing through coordinates**** = (6, 4)**

**The equation is symmetric as considered with y – axis.**

**Sol:**

**Given:**

**(i) Vertex is at origin i.e., (0, 0)**

**(ii) Passing through coordinates**** = (6, 4)**

As the vertex of a parabola lies at origin and the passing through coordinates = (6, 4).

The required equation is either x^{2} = 4ay or x^{2} = – 4ay

**The coordinates (6, 4) is in first quadrant.**

Therefore, the equation of the parabola = x^{2} = 4ay, whereas (6, 4) should satisfy the equation

So,

6^{2} = 4a (4)

36 = 16 a

a = \(\frac{36}{16}\)

a = \(\frac{9}{4}\)

**The parabola’s equation is:**

x^{2} = 4(\(\frac{9}{4}\))y

**x**^{2}** = 9 y**

**Exercise 11.3**

** ****Q.1: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse \(\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1\).**

** **

**Sol:**

**Given:**

\(\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1\) **. . . . . . . . (1) is the equation of the ellipse.**

As we know that the **denominator** of \(\frac{x^{2}}{25}\) > the denominator of \(\frac{y^{2}}{9}\)

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . (2)**

**Now, on comparing equation (1) and equation (2) we will get:**

We get:

**m = 5, n = 3**

The **vertices** **coordinates** are (m, 0) and (- m, 0) **= (5, 0), (- 5, 0)**

The **foci’s** **coordinates** are (c, 0) and (- c, 0) **= (4, 0) and (- 4, 0)**

Length of the axis:

**Major axis** = 2m **= 10**

**Minor axis** = 2n **= 6**

**Length of the latus rectum =** \(\frac{2n^{2}}{m}\) = \(\frac{2.3^{2}}{5} = \frac{18}{5}\)

**Eccentricity, e =** \(\frac{c}{m}\) = \(\frac{4}{5}\)

**Q.2: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse \(\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1\).**

** **

**Sol:**

**Given:**

\(\frac{x^{2}}{9} + \frac{y^{2}}{16} = 1\) **. . . . . . . . . . (1) is the equation of the ellipse.**

As we know that the **denominator** of \(\frac{x^{2}}{9}\) < the denominator of \(\frac{y^{2}}{16}\)

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

\(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) **. . . . . . . . . . . . (2)**

**Now, on comparing equation (1) and equation (2) we will get:**

We get,:

**m = 4, n = 3**

The **vertices coordinates** are (0, m) and (0, – m) **= (0, 4), (0, – 4)**

The **foci’s coordinates** are (0, c) and (0, – c)** = (\(\sqrt{7}\), 0) and (-\(\sqrt{7}\), 0)**

**Length of the axis:**

**Major axis** = 2m **= 8**

**Minor axis** = 2n **= 6**

**Length of the latus rectum =** \(\frac{2n^{2}}{m}\) = \(\frac{2.3^{2}}{4} = \frac{9}{2}\)

**Eccentricity, e =** \(\frac{c}{m}\) = \(\frac{\sqrt{7}}{4}\)

**Q.3: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse \(\frac{x^{2}}{25} + \frac{y^{2}}{4} = 1\).**

** **

**Sol:**

**Given:**

\(\frac{x^{2}}{25} + \frac{y^{2}}{4} = 1\) . . . . . . . . . . **(1) is the equation of the ellipse.**

As we know that the **denominator** of \(\frac{x^{2}}{25}\) > the denominator of \(\frac{y^{2}}{4}\)

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . (2)**

**Now, on comparing equation (1) and equation (2) we will get:**

We get:

**m = 5, n = 2**

The **vertices** **coordinates** are (m, 0) and (- m, 0) **= (5, 0), (- 5, 0)**

The **foci’s coordinates** are (c, 0) and (- c, 0) **= (\(\sqrt{21}\), 0) and (-\(\sqrt{21}\), 0)**

**Length of the axis:**

**Major axis =** 2m = **10**

**Minor axis =** 2n = **4**

**Length of the latus rectum =** \(\frac{2n^{2}}{m}=\frac{2.2^{2}}{5} = \frac{8}{5}\)

**Eccentricity, e =** \(\frac{c}{m}\) = \(\frac{\sqrt{21}}{5}\)

**Q.4: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse \(\frac{x^{2}}{36} + \frac{y^{2}}{121} = 1\).**

** **

**Sol:**

**Given:**

\(\frac{x^{2}}{36} + \frac{y^{2}}{121} = 1\) **. . . . . . . . . . (1) is the equation of the ellipse.**

As we know that the **denominator** of \(\frac{x^{2}}{36}\) < the denominator of \(\frac{y^{2}}{121}\)

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

\(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) **. . . . . . . . . . . . . . . (2)**

**Now, on comparing equation (1) and equation (2) we will get:**

We get:

**m = 11, n = 6**

The **vertices coordinates** are (0, m) and (0, – m) **= (0, 11), (0, – 11)**

The **foci’s coordinates** are (0, c) and (0, – c) **= (0, \(\sqrt{85}\)) and (0, –\(\sqrt{85}\))**

**Length of the axis:**

**Major axis =** 2m **= 22**

**Minor axis =** 2n **= 12**

Length of the **latus rectum** = \(\frac{2n^{2}}{m}\) = \(\frac{2.6^{2}}{11} = \frac{72}{11}\)

**Eccentricity, e =** \(\frac{c}{m}\) **= \(\frac{\sqrt{85}}{11}\)**

**Q.5: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse \(\frac{x^{2}}{64} + \frac{y^{2}}{25} = 1\).**

** **

**Sol:**

**Given:**

\(\frac{x^{2}}{64} + \frac{y^{2}}{25} = 1\) **. . . . . . . . . . . . . (1) is the equation of the ellipse.**

As we know that the **denominator** of \(\frac{x^{2}}{64}\) > the denominator of \(\frac{y^{2}}{25}\)

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . (2)**

**Now, on comparing equation (1) and equation (2) we will get:**

We get:

**m = 8, n = 5**

The **vertices coordinates** are (m, 0) and (- m, 0) **= (8, 0), (- 8, 0)**

The **foci’s coordinates **are (c, 0) and (- c, 0) **= (\(\sqrt{39}\), 0) and (-\(\sqrt{39}\), 0)**

**Length of the axis:**

**Major axis** = 2m **= 16**

**Minor axis** = 2n **= 10**

Length of the **latus rectum** = \(\frac{2n^{2}}{m}\) = \(\frac{2.5^{2}}{8} = \frac{25}{4}\)

**Eccentricity, e** = \(\frac{c}{m}\) **= \(\frac{\sqrt{39}}{8}\)**

**Q.6: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse \(\frac{x^{2}}{400} + \frac{y^{2}}{100} = 1\).**

** **

**Sol:**

**Given:**

\(\frac{x^{2}}{400} + \frac{y^{2}}{100} = 1\)** . . . . . . . . . . . . (1) is the equation of the ellipse.**

As we know that the **denominator** of \(\frac{x^{2}}{400}\) > the denominator of \(\frac{y^{2}}{100}\)

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . (2)**

**Now, on comparing equation (1) and equation (2) we will get:**

We get:

**m = 20, n = 10**

The **vertices coordinates** are (m, 0) and (- m, 0) **= (20, 0), (- 20, 0)**

The **foci’s coordinates** are (c, 0) and (- c, 0) **= (\(10 \sqrt{3}\), 0) and (-\(10 \sqrt{3}\), 0)**

**Length of the axis:**

**Major axis** = 2m **= 40**

**Minor axis** = 2n **= 20**

**Length of the latus rectum** = \(\frac{2n^{2}}{m}\) = \(\frac{2.10^{2}}{20} = 10\)

**Eccentricity, e** = \(\frac{c}{m}\) = \(\frac{10 \sqrt{3}}{20} = \frac{\sqrt{3}}{2}\)

**Q.7: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 49x ^{2} + 9y^{2 }= 441**

** **

**Sol:**

**Given:**

**49x ^{2} + 9y^{2 }= 441**

\(\frac{x^{2}}{9} + \frac{y^{2}}{49} = 1\) ** . . . . . . . . . . . . . . . (1) is the equation of the ellipse.**

As we know that the **denominator** of \(\frac{x^{2}}{9}\) < the denominator of \(\frac{y^{2}}{49}\)

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

\(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) **. . . . . . . . . . . . . . . (2)**

**Now, on comparing equation (1) and equation (2) we will get:**

We get:

**m = 7, n = 3**

The **vertices coordinates** are (0, m) and (0, – m) **= (0, 7), (0, – 7)**

The **foci’s coordinates** are (0, c) and (0, – c) **= (0, \(2 \sqrt{10}\)) and (0, \(2 \sqrt{10}\))**

**Length of the axis:**

**Major axis **= 2m** = 14**

**Minor axis** = 2n **= 6**

**Length of the** **latus rectum =** \(\frac{2n^{2}}{m}\) = \(\frac{2.3^{2}}{7} = \frac{18}{7}\)

**Eccentricity, e =** \(\frac{c}{m}\) = \(\frac{2 \sqrt{10}}{7}\)

**Q.8: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 4x ^{2} + 16y^{2 }= 64**

** **

**Sol:**

**Given:**

**4x ^{2} + 16y^{2 }= 64**

\(\frac{x^{2}}{16} + \frac{y^{2}}{4} = 169\)** . . . . . . . . . . . . . . . . . (1) is the equation of the ellipse.**

As we know that the **denominator** of \(\frac{x^{2}}{16}\) > the denominator of \(\frac{y^{2}}{4}\)

So, along x – axis consist the major axis and along the y – axis consists the minor axis.

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\)** . . . . . . . . . . . . . . . (2)**

**Now, on comparing equation (1) and equation (2) we will get:**

We get:

**m = 4, n = 2**

The **vertices coordinates** are (m, 0) and (- m, 0) **= (4, 0), (- 4, 0)**

The **foci’s coordinates** are (c, 0) and (- c, 0) **= (\(2 \sqrt{3}\), 0) and (-\(2 \sqrt{3}\), 0)**

**Length of the axis:**

**Major axis** = 2m **= 8**

**Minor axis** = 2n **= 4**

**Length of the latus rectum =** \(\frac{2n^{2}}{m}\) = \(\frac{2.4^{2}}{4} = 8\)

**Eccentricity, e =** \(\frac{c}{m}\) = \(\frac{2 \sqrt{3}}{4} = \frac{\sqrt{3}}{2}\)

**Q.9: Obtain the coordinates of the vertices, the foci, the major and the minor axis, the length of the latus rectum and the eccentricity of the ellipse 4x ^{2} + y^{2 }= 4**

** **

**Sol:**

Given:

4x^{2} + y^{2 }= 4

\(\frac{x^{2}}{1} + \frac{y^{2}}{4} = 1\) **. . . . . . . . . . . . . . . . (1) is the equation of the ellipse.**

As we know that the **denominator** of \(\frac{x^{2}}{1}\) < the denominator of \(\frac{y^{2}}{4}\)

So, along y – axis consist the major axis and along the x – axis consists the minor axis.

\(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) **. . . . . . . . . . . . . (2)**

**Now, on comparing equation (1) and equation (2) we will get:**

We get:

**m = 2, n = 1**

The **vertices coordinates** are (0, m) and (0, – m) **= (0, 2), (0, – 2)**

The **foci’s coordinates** are (0, c) and (0, – c) **= (0, \(\sqrt{3}\)) and (0, –\( \sqrt{3}\))**

Length of the axis:

**Major axis** = 2m **= 4**

**Minor axis** = 2n **= 2**

**Length of the latus rectum =** \(\frac{2n^{2}}{m}\) = \(\frac{2.1^{2}}{2} = 1\)

**Eccentricity, e =** \(\frac{c}{m}\) = \(\frac{\sqrt{3}}{2}\)

**Q.10: For the given condition obtain the equation of the ellipse.**

**(i) Vertices (****±****6, 0)**

**(ii) Foci (****±****3, 0)**

** **

**Sol:**

**Given:**

**(i) Vertices (****±****6, 0)**

**(ii) Foci (****±****3, 0)**

The vertices are represented along x – axis.

The required **equation of the ellipse **is of the form \(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . . . (1)**

We get:

**m = 6 (semi major axis), c = 3**

As we know:

**m ^{2} = n^{2} + c^{2}**

6^{2} = n^{2} + 3^{2}

n^{2} = 6^{2} – 3^{2}

n^{2} = 36 – 9 = 27

**n =** \(\sqrt{27} = 3\sqrt{3}\)

**Hence, \(\frac{x^{2}}{36} + \frac{y^{2}}{27} = 1\) is the equation of the ellipse.**

**Q.11: For the given condition obtain the equation of the ellipse.**

**(i) Vertices (0, ****±****8)**

**(ii) Foci ( 0, ****±****4)**

** **

**Sol:**

**Given:**

**(i) Vertices (0, ****±****8)**

**(ii) Foci ( 0, ****±****4)**

The vertices are represented along y – axis.

The required **equation of the ellipse** is of the form \(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) **. . . . . . . . . . . . . . .(1)**

We get:

**m = 8 (semi major axis), c = 4**

As we know:

**m ^{2} = n^{2} + c^{2}**

8^{2} = n^{2} + 4^{2}

n^{2} = 8^{2} – 4^{2}

n^{2} = 64 – 16 = 48

**n =** \(\sqrt{48} = 4\sqrt{3}\)

**Hence, \(\frac{x^{2}}{48} + \frac{y^{2}}{64} = 1\) is the equation of the ellipse.**

** **

**Q.12: For the given condition obtain the equation of the ellipse.**

**(i) Vertices (****±****5, 0)**

**(ii) Foci (****±****3, 0)**

** **

**Sol:**

**Given:**

**(i) Vertices (****±**** 5, 0)**

**(ii) Foci (****±**** 3, 0)**

The vertices are represented along x – axis.

The required **equation of the ellipse **is of the form \(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . (1)**

We get:

**m = 5 (semi major axis), c = 3**

As we know:

**m ^{2} = n^{2} + c^{2}**

5^{2} = n^{2} + 3^{2}

n^{2} = 5^{2} – 3^{2}

n^{2} = 25 – 9 = 16

**n =** \(\sqrt{16} = 4\)

**Hence, \(\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1\) is the equation of the ellipse.**

** **

**Q.13: For the given condition obtain the equation of the ellipse.**

**(i) Coordinates of major axis (****±****5, 0)**

**(ii) Coordinates of minor axis (0, ****±****3)**

** **

**Sol:**

**Given:**

**(i) Coordinates of major axis (****±****5, 0)**

**(ii) Coordinates of minor axis (0, ****±****3)**

The major axis is represented along x – axis.

The required **equation of the ellipse** is of the form

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . (1)**

**We get:**

**m = 5 (semi major axis), n = 3**

**Hence, \(\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1\) is the equation of the ellipse.**

** **

**Q.14: For the given condition obtain the equation of the ellipse.**

**(i) Coordinates of major axis (0, ****±****2)**

**(ii) Coordinates of minor axis (****±****\(\sqrt{3}\), 0)**

** **

**Sol:**

**Given:**

**(i) Coordinates of major axis (0, ****±****2)**

**(ii) Coordinates of minor axis (****±****\(\sqrt{3}\), 0)**

The major axis is represented along y – axis.

The required **equation of the ellipse** is of the form \(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) **. . . . . . . . . . . . . . . . . (1)**

We get:

**m = 2 (semi major axis), n = \(\sqrt{3}\)**

**Hence, \(\frac{x^{2}}{3} + \frac{y^{2}}{4} = 1\) is the equation of the ellipse.**

** **

**Q.15: For the given condition obtain the equation of the ellipse.**

**(i) Length of major axis 30**

**(ii) Coordinates of foci (****±****4, 0)**

** **

**Sol:**

**Given:**

**(i) Length of major axis 30**

**(ii) Coordinates of foci (****±****4, 0)**

The major axis are represented along x – axis as foci is along x – axis

The required **equation of the ellipse** is of the form \(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . . . . . (1)**

We get:

2m = 30 (semi major axis),

**m = 15**

**c = 4**

As we know:

** m ^{2} = n^{2} + c^{2}**

15^{2} = n^{2} + 4^{2}

n^{2} = 15^{2} – 4^{2}

n^{2} = 225 – 16 = 209

**n =** \(\sqrt{209}\)

**Hence, \(\frac{x^{2}}{225} + \frac{y^{2}}{209} = 1\) is the equation of the ellipse.**

**Q.16: For the given condition obtain the equation of the ellipse.**

**(i) Length of minor axis 26**

**(ii) Coordinates of foci (0, ****±****9)**

** **

**Sol:**

**Given:**

**(i) Length of minor axis 26**

**(ii) Coordinates of foci (0, ****±****9)**

The major axis are represented along y – axis as foci is along y – axis

The required **equation of the ellipse** is of the form

\(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) **. . . . . . . . . . . . . (1)**

We get:

2n = 26 (semi major axis),

**n = 13**

**c = 9**

As we know:

**m ^{2} = n^{2} + c^{2}**

m^{2} = 13^{2} + 9^{2}

m^{2} = 169 + 81

m^{2} = 250

**m =** \(\sqrt{250}\) = \(5 \sqrt{10}\)

**Hence, \(\frac{x^{2}}{169} + \frac{y^{2}}{250} = 1\) is the equation of the ellipse.**

** **

**Q.17: For the given condition obtain the equation of the ellipse.**

**(i) Coordinates of foci (****±****4, 0)**

**(ii) m = 6**

** **

**Sol:**

**Given:**

**(i) Coordinates of foci (****±****4, 0)**

**(ii) m = 6**

The major axis are represented along x – axis as foci is along x – axis

The required **equation of the ellipse** is of the form

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . (1)**

We get:

**c = 4**

**m = 6**

As we know:

**m ^{2} = n^{2} + c^{2}**

6^{2} = n^{2} + 4^{2}

n^{2} = 36 – 16

n^{2} = 20

**n =** \(\sqrt{20}\) = \(2 \sqrt{5}\)

**Hence, \(\frac{x^{2}}{36} + \frac{y^{2}}{20} = 1\) is the equation of the ellipse.**

**Q.18: For the given condition obtain the equation of the ellipse.**

**(i) n = 2, c = 3 (on the x – axis)**

**(ii) Coordinates of centre (0, 0)**

** **

**Sol:**

**Given:**

**(i) n = 2, c = 3 (on the x – axis)**

**(ii) Coordinates of centre (0, 0) i.e., centre is at origin.**

The major axis are represented along x – axis as foci is along x – axis

The required **equation of the ellipse** is of the form:

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . (1)**

We have:

**n = 2, c = 3 **

As we know:

**m ^{2} = n^{2} + c^{2}**

m^{2} = 2^{2} + 3^{2}

m^{2} = 4 + 9

m^{2} = 13

**m =** \(\sqrt{13}\)

**Hence, \(\frac{x^{2}}{13} + \frac{y^{2}}{4} = 1\) is the equation of the ellipse.**

** **

**Q.19: For the given condition obtain the equation of the ellipse.**

**(i) Centre is at origin and major axis is along the y – axis**

**(ii) It passes through the points (1, 6) and (3, 2)**

** **

**Sol:**

**Given:**

**(i) Centre is at origin and major axis is along the y – axis**

**(ii) It passes through the points (1, 6) and (3, 2)**

The major axis are represented along y – axis as foci is along y – axis

The required **equation of the ellipse** is of the form:

\(\frac{x^{2}}{n^{2}} + \frac{y^{2}}{m^{2}} = 1\) **. . . . . . . . . . . . . . . (1)**

As the ellipse passes through (2, 1) and (2, 3) points, then

\(\frac{1^{2}}{n^{2}} + \frac{6^{2}}{m^{2}} = 1\) \(\frac{3^{2}}{n^{2}} + \frac{2^{2}}{m^{2}} = 1\)\(\frac{1}{n^{2}} + \frac{36}{m^{2}}\) = 1 **. . . . . . . . . . . . . . . (a)**

\(\frac{9}{n^{2}} + \frac{4}{m^{2}}\) = 1 **. . . . . . . . . . . . . . . . (b)**

**Now, on comparing equation (a) and equation (b) we will get:**

**m ^{2} = 40, n^{2} = 10**

**Hence, \(\frac{y^{2}}{40} + \frac{x^{2}}{10} = 1\) is the equation of the ellipse.**

**Q.20: For the given condition obtain the equation of the ellipse.**

**(i) Major axis is represented along x – axis**

**(ii) It passes through the points (6, 2) and (4, 3)**

** **

**Sol:**

**Given:**

**(i) Major axis is represented along x – axis**

**(ii) It passes through the points (6, 2) and (4, 3)**

The major axis are represented along y – axis as foci is along y – axis

The required **equation of the ellipse** is of the form:

\(\frac{x^{2}}{m^{2}} + \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . . (1)**

As the ellipse passes through (6, 2) and (4, 3) points, then

\(\frac{6^{2}}{m^{2}} + \frac{2^{2}}{n^{2}} = 1\) \(\frac{4^{2}}{m^{2}} + \frac{3^{2}}{n^{2}} = 1\)\(\frac{36}{m^{2}} + \frac{4}{n^{2}}\) = 1 **. . . . . . . . . . . . . . . . (a)**

\(\frac{16}{m^{2}} + \frac{9}{n^{2}}\) = 1 **. . . . . . . . . . . . . . . . (b)**

**Now, on comparing equation (a) and equation (b) we will get:**

**m ^{2} = 52 and n^{2} = 13**

**Hence, \(\frac{y^{2}}{13} + \frac{x^{2}}{52} = 1\) is the equation of the ellipse.**

**Exercise 11.4**

**Q.1: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola \(\frac{x^{2}}{25} – \frac{y^{2}}{16} = 1\).**

** **

**Sol:**

**Given:**

\(\frac{x^{2}}{25} – \frac{y^{2}}{16} = 1\) **. . . . . . . . . . . . . (1) is the equation of the hyperbola.**

The **specific equation for hyperbola** is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . . . .(2)**

**Now, on comparing equation (1) and equation (2) we will get:**

We get:

**m = 5 and n = 4**

The **vertices coordinates** are (m, 0) and (- m, 0) **= (5, 0), (- 5, 0)**

The **foci’s coordinates** are (c, 0) and (- c, 0) **= (\(\sqrt{41}\), 0) and (-\(\sqrt{41}\), 0)**

**Length of the latus rectum =** \(\frac{2n^{2}}{m}\) = \(\frac{2.4^{2}}{5} = \frac{32}{5}\)

**Eccentricity, e =** \(\frac{c}{m}\) = \(\frac{\sqrt{41}}{5}\)

**Q.2: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola \(\frac{x^{2}}{1} – \frac{y^{2}}{8} = 1\).**

** **

**Sol:**

**Given:**

\(\frac{x^{2}}{1} – \frac{y^{2}}{8} = 1\) **. . . . . . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.**

The **specific equation for hyperbola** is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . . . . . . . . .(2)**

**Now, on comparing equation (1) and equation (2) we will get:**

**m =** 1and **n =** \(\sqrt{8}\)

As we know:

\(c = \sqrt{m^{2} + n^{2}} = \sqrt{1 + 8} = \sqrt{9}\) = 3

The **vertices coordinates** are (0, m) and (0, – m) **= (0, 1), (0, – 1)**

The **foci’s coordinates** are (0, c) and (0, – c) **= (0, 3), (0, – 3)**

**Length of the latus rectum =** \(\frac{2n^{2}}{m}\) = \(\frac{16}{1} = 16\)

**Eccentricity, e =** \(\frac{c}{m}\) = \(\frac{\sqrt{3}}{1}\) = 3

**Q.3: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 25 y ^{2} – 4 x^{2} = 100**

** **

**Sol:**

**Given:**

**25 y ^{2} – 4 x^{2} = 100**

\(\frac{y^{2}}{4} – \frac{x^{2}}{25} = 1\) **. . . . . . . . . . . . . . . (1) is the equation of the hyperbola.**

The **specific equation for hyperbola** is \(\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . .(2)

**Now, on comparing equation (1) and equation (2) we will get:**

**m = 2** and **n = 5.**

The **vertices coordinates** are (0, m) and (0, – m) **= (0, 2) and (0, – 2)**

The **foci’s coordinates** are (0, c) and (0, – c) = (0, \(\sqrt{29}\)) and (0, –\(\sqrt{29}\))

**Length of the latus rectum =** \(\frac{2.n^{2}}{m}\) = \(\frac{2.5^{2}}{2} = \frac{50}{2}\) = 25

**Eccentricity, e =** \(\frac{c}{m}\) = \(\frac{\sqrt{29}}{2}\)

**Q.4: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 9 x ^{2} – 4 y^{2} = 36**

** **

**Sol:**

**Given:**

**9 x ^{2} – 4 y^{2} = 36**

\(\frac{x^{2}}{4} – \frac{y^{2}}{9} = 1\)** . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.**

The **specific equation for hyperbola** is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . . (2)**

**Now, on comparing equation (1) and equation (2) we will get:**

**m = 2** and **n = 3.**

The **vertices coordinates** are (m, 0) and (- m, 0) **= (2, 0), (- 2, 0)**

The **foci’s coordinates** are (c, 0) and (- c, 0) = (\(\sqrt{13}\), 0) and (-\(\sqrt{13}\), 0)

**Length of the latus rectum =** \(\frac{2n^{2}}{m}\) = \(\frac{2.3^{2}}{2} = 9\)

**Eccentricity, e =** \(\frac{c}{m}\) = \(\frac{\sqrt{13}}{2}\)

**Q.5: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 6 y ^{2} – 16 x^{2} = 96**

** **

**Sol:**

**Given:**

**6 y ^{2} – 16 x^{2} = 96**

\(\frac{y^{2}}{16} – \frac{x^{2}}{6} = 1\)** . . . . . . . . . . . . . . (1) is the equation of the hyperbola.**

The **specific equation for hyperbola** is \(\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . . . .(2)**

**Now, on comparing equation (1) and equation (2) we will get:**

**m =** 4 and **n = **\(\sqrt{6}\)

The **vertices coordinates** are (0, m) and (0, – m) **= (0, 4) and (0, – 4)**

The **foci’s coordinates** are (0, c) and (0, – c) **= (0, \(\sqrt{10}\)) and (0, –\(\sqrt{10}\))**

**Length of the latus rectum =** \(\frac{2.n^{2}}{m}\) = \(\frac{2. \sqrt{6}^{2}}{4} = \frac{12}{4}\) = 3

**Eccentricity, e =** \(\frac{c}{m}\) = \(\frac{\sqrt{10}}{4}\)

**Q.6: Obtain the coordinates of the vertices, the foci, the length of the latus rectum and the eccentricity of the hyperbola 64 y ^{2} – 9 x^{2} = 96**

** **

**Sol:**

**Given:**

**6 y ^{2} – 16 x^{2} = 96**

\(\frac{y^{2}}{16} – \frac{x^{2}}{6} = 1\)** . . . . . . . . . . . . . . . . (1) is the equation of the hyperbola.**

The specific equation for hyperbola is \(\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1\) . . . . . . . . . . . . . . . . . . . . (2)

**Now, on comparing equation (1) and equation (2) we will get:**

**m =** 4 and **n =** \(\sqrt{6}\)

The **vertices** **coordinates** are (0, m) and (0, – m) **= (0, 4) and (0, – 4)**

The **foci’s coordinates** are (0, c) and (0, – c) **= (0, \(\sqrt{10}\)) and (0, –\(\sqrt{10}\))**

**Length of the latus rectum =** \(\frac{2.n^{2}}{m}\) = \(\frac{2. \sqrt{6}^{2}}{4} = \frac{12}{4}\) = 3

**Eccentricity, e =** \(\frac{c}{m}\) = \(\frac{\sqrt{10}}{4}\)

**Q.7: For the given condition obtain the equation of the hyperbola.**

**(i) Coordinates of the vertices (****±****3, 0)**

**(ii) Coordinates of the foci (****±****4, 0)**

** **

**Sol:**

**Given:**

**(i) Coordinates of the vertices (****±****3, 0)**

**(ii) Coordinates of the foci (****±****4, 0)**

The **specific equation for hyperbola** is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . . (1)**

As, **coordinates of the vertices** (±3, 0), **m = 3**

And, **coordinates of the foci** (±4, 0), **c = 4**

We know that:

**c ^{2} = m^{2} + n^{2}**

4^{2} = 3^{2} + n^{2}

n^{2} = 16 – 9 = 5

**\(\frac{x^{2}}{9} – \frac{y^{2}}{5} = 1\) is the equation of the hyperbola.**

**Q.8: For the given condition obtain the equation of the hyperbola.**

** (i) Coordinates of the vertices (0, ****±****7)**

**(ii) Coordinates of the foci (0, ****±****9)**

** **

**Sol:**

**Given:**

**(i) Coordinates of the vertices (0, ****±****7)**

**(ii) Coordinates of the foci (0, ****±****9)**

The **specific equation for hyperbola** is \(\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . . . . . (1)**

As, **coordinates of the vertices** (0, ±7), **m = 7**

And, **coordinates of the foci** (0, ±9), **c = 9**

We know that:

**c ^{2} = m^{2} + n^{2}**

9^{2} = 7^{2} + n^{2}

**n ^{2} =** 81 – 49

**= 32**

**\(\frac{y^{2}}{49} – \frac{x^{2}}{32} = 1\) is the equation of the hyperbola.**

**Q.9: For the given condition obtain the equation of the hyperbola.**

** (i) Coordinates of the vertices (0, ****±****8)**

**(ii) Coordinates of the foci (0, ****±****11)**

** **

**Sol:**

**Given:**

**(i) Coordinates of the vertices (0, ****±****8)**

**(ii) Coordinates of the foci (0, ****±****11)**

The **specific equation for hyperbola** is \(\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . . . . . . . . (1)**

As, **coordinates of the vertices** (0, ±8), **m = 8**

And, **coordinates of the foci** (0, ±11), **c = 11**

We know that:

**c ^{2} = m^{2} + n^{2}**

11^{2} = 8^{2} + n^{2}

**n ^{2} =** 121 – 64

**= 57**

**\(\frac{y^{2}}{64} – \frac{x^{2}}{57} = 1\) is the equation of the hyperbola.**

**Q.10: For the given condition obtain the equation of the hyperbola.**

** (i) Coordinates of the foci (****±****7, 0), **

**(ii) The length of the transverse axis is 12**

** **

**Sol:**

**Given:**

**(i) Coordinates of the foci (****±****7, 0),**

**(ii) The length of the transverse axis is 12**

The **specific equation for hyperbola** is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . . . . (1)**

And**, coordinates of the foci** (±7, 0), **c = 7**

As, the **transverse axis is of length 12**,

2 m = 12

**m = 6**

We know that:

**c ^{2} = m^{2} + n^{2}**

7^{2} = 6^{2} + n^{2}

**n ^{2} =** 49 – 36

**= 13**

**\(\frac{x^{2}}{36} – \frac{y^{2}}{13} = 1\) is the equation of the hyperbola.**

**Q.11: For the given condition obtain the equation of the hyperbola.**

**(i) Coordinates of the foci (0, ****±****8)**

**(ii) The length of the conjugate axis is 16**

** **

**Sol:**

**Given:**

**(i) Coordinates of the foci (0, ****±****9)**

**(ii) The length of the conjugate axis is 16**

The **specific equation for hyperbola** is \(\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . . (1)**

And, **coordinates of the foci** (0, ±9), **c = 9**

As, the **conjugate axis is of length 16,**

2 m = 16

**m = 8**

**Therefore, m ^{2} = 64**

We know that:

**c ^{2} = m^{2} + n^{2}**

9^{2} = 8^{2} + n^{2}

**n ^{2} **= 81 – 64

**= 17**

**\(\frac{y^{2}}{64} – \frac{x^{2}}{17} = 1\) is the equation of the hyperbola.**

**Q.12: For the given condition obtain the equation of the hyperbola.**

**(i) Coordinates of the foci (****±****\(2\sqrt{3}\), 0 )**

**(ii) The length of the latus rectum is 8**

** **

**Sol:**

**Given:**

**(i) Coordinates of the foci (****±****\(2\sqrt{3}\), 0 )**

**(ii) The length of the latus rectum is 8**

The **specific equation for hyperbola** is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . (1)**

And, the **coordinates of the foci** (±\(2\sqrt{3}\), 0 ), **c = \(2\sqrt{3}\)**

As, the **latus rectum is of length 8**

We know that:

**m ^{2} + n^{2} = c^{2}**

m^{2} + 4m = \((2\sqrt{3})^{2}\)

m^{2} + 4m = 12

m^{2} + 4m – 12 = 0

m^{2 }+ 6m – 2m – 12 = 0

m (m + 6) – 2 (m + 6) = 0

m = – 6 and m = 2

**m is non – negative** so, **m = 2** and **n ^{2} = 8 [Since, n^{2} = 12 – 2^{2} = 8] **

**\(\frac{x^{2}}{4} – \frac{y^{2}}{8} = 1\) is the equation of the hyperbola.**

**Q.13: For the given condition obtain the equation of the hyperbola.**

**(i) Coordinates of the foci (****±****\(2\sqrt{6}\), 0)**

**(ii) The length of the latus rectum is 10**

** **

**Sol:**

**Given:**

**(i) Coordinates of the foci (****±****\(2\sqrt{3}\), 0 )**

**(ii) The length of the latus rectum is 10**

The **specific equation for hyperbola** is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . . . (1)**

And, coordinates of the foci (±\(2\sqrt{6}\), 0 ), **c = \(2\sqrt{6}\)**

As, the **latus rectum is of length 10**

Therefore, \(\frac{2n^{2}}{m} = 10 \\ 2n^{2} = 10 m \\ n^{2} = 5m\)

We know that:

** m ^{2} + n^{2} = c^{2}**

m^{2} + 5 m = \((2\sqrt{6})^{2}\)

m^{2} + 5 m = 24

m^{2} + 5 m – 24 = 0

m^{2 }+ 8 m – 3 m – 24 = 0

m (m + 8) – 3 (m + 8) = 0

m = – 8 and **m = 3**

**m is non – negative so, m =** 3 and **n ^{2} =** 5m

**= 15**

**\(\frac{x^{2}}{9} – \frac{y^{2}}{15} = 1\) is the equation of the hyperbola.**

** **

**Q.14: For the given condition obtain the equation of the hyperbola.**

**(i) Coordinates of the vertices (****±****9, 0)**

**(ii) Eccentricity (e) = \(\frac{9}{4}\)**

** **

**Sol:**

**Given:**

** (i) Coordinates of the vertices (****±****9, 0)**

**(ii) Eccentricity (e) = \(\frac{9}{4}\)**

The **specific equation for hyperbola** is \(\frac{x^{2}}{m^{2}} – \frac{y^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . . . . . . (1)**

And, **Coordinates of the vertices** (±9, 0),** m = 9**

As, **the eccentricity (e) =** \(\frac{9}{4}\),

**c =** \(\frac{81}{4}\)

**We know that:**

** m ^{2} + n^{2} = c^{2}**

9^{2} + n^{2} = \((\frac{81}{4})^{2}\)

n^{2} = \((\frac{81}{4})^{2}\) – 9^{2}

n^{2 }= \(\frac{6561}{16} – 81\)

n^{2 }= \(\frac{6561 – 1296}{16} = \frac{5265}{16}\)

**\(\frac{x^{2}}{81} – \frac{16 y^{2}}{5265} = 1\) is the equation of the hyperbola.**

** **

**Q.15: For the given condition obtain the equation of the hyperbola.**

**(i) Coordinates of the foci (0, ****±**** \(\sqrt{10}\))**

**(ii) It passes through (2, 3)**

** **

**Sol:**

**Given:**

**(i) Coordinates of the foci (0, ****±****\(\sqrt{10}\))**

**(ii) It passes through (2, 3)**

The **specific** **equation for hyperbola** is \(\frac{y^{2}}{m^{2}} – \frac{x^{2}}{n^{2}} = 1\) **. . . . . . . . . . . . (1)**

And,** coordinates of the foci **(0, ±\(\sqrt{10}\)), c = \(\sqrt{10}\)

We know that:

** m ^{2} + n^{2} = c^{2}**

n^{2} = c^{2} – m^{2}

n^{2} = 10 – m^{2} **. . . . . . . . . . . . (2)**

**As the hyperbola passes through (2, 3)**

\(\frac{9}{m^{2}} – \frac{4}{n^{2}} = 1\) **. . . . . . . . . . . (3)**

**Putting Equation (2) in equation (3), we get:**

9 (10 – m^{2}) – 4m^{2} = m^{2} (10 – m^{2})

m^{4} – 23m^{2} + 90 = 0

m^{4} – 18m^{2} – 5m^{2} + 90 = 0

(m^{2} – 5) (m^{2} – 18) = 0

**m ^{2 }= 5 or 18**

As in hyperbola:

m^{2} < c^{2}

**m ^{2 }= 5**

**n ^{2 }=** 10 – 5 [From Equation (2)]

**= 5**

**\(\frac{y^{2}}{5} – \frac{x^{2}}{5} = 1\) is the equation of the hyperbola.**