NCERT Exemplar Solutions for Class 11 Maths Chapter 11 Conic Sections is an important study resource available for students. NCERT Exemplar books provide multiple-choice questions and very short answer types of questions, by which students can enhance their analytical thinking and time management skills. They have proven to be essential for learning the syllabus and building the confidence that is required to face their exams. These questions help the students in understanding the types of questions that will be asked in the exams. Solutions for all Conic Sections Exercise problems help students in preparing for the exams in a better way. The NCERT Exemplar solutions available at BYJUâ€™S will help students to ace their exams by preparing them well in advance.

Chapter 11 of NCERT Exemplar Solutions for Class 11 Maths Conic Sections explains parabola, circles and section of a cone. The exercise-wise solutions of these topics are available here in PDF format, which can be easily downloaded by the students. These solutions are prepared by subject experts at BYJUâ€™S in order to guide them in their academics. Some of the essential topics of this chapter are listed below.

- Sections of a cone
- Circle and equation of a circle
- Parabola
- Directrix of the parabola
- Standard equations of parabola
- Focal distance of a point of parabola
- Ellipse
- Focal Distance of a ellipse
- Hyperbola
- Focal distance of hyperbola
- Parametric equation of conics

## Download the PDF of NCERT Exemplar Solutions For Class 11 Maths Chapter 11 Conic Sections

### Access Answers to NCERT Exemplar Solutions For Class 11 Maths Chapter 11 – Conic Sections

Exercise Page No: 202

Short Answer type:

**1. Find the equation of the circle which touches the both axes in first quadrant and whose radius is a.**

**Solution:**

The circle touches both the x and y axes in the first quadrant and the radius is a.

For a circle of radius a, the centre is (a, a).

The equation of a circle having centre (h, k), having radius as r units, is

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

Therefore, the equation of the circle becomes (x â€“ a)^{2}Â + (y â€“ a)^{2}Â = a^{2}

x^{2}Â – 2ax + a^{2}Â + y^{2}Â – 2ay + a^{2}Â – a^{2}Â = 0

x^{2}Â – 2ax + y^{2}Â – 2ay + a^{2}Â = 0

**Solution:**

**3. If a circle passes through the point (0, 0) (a, 0), (0, b) then find the coordinates of its centre.**

**Solution:**

The equation of a circle having centre (h, k), having radius as r units, is

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

Putting the values of given co-ordinates in the above expression,

(0, 0)

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

h^{2}Â + k^{2}Â = r^{2}

(a, 0)

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

a^{2}Â + h^{2}Â – 2ah + k^{2}Â = r^{2}Â —— (1)

(0, b)

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

h^{2}Â + b^{2}Â + k^{2}Â – 2bk = r^{2}Â ——- (2)

On solving equations (1) & (2), respectively,

a (a â€“ 2h) = 0

b (b â€“ 2k) = 0

So, a = 0 or 2h

b = 0 or 2k respectively.

Since the circle passes through the centre (0, 0), so the co-ordinates are

a = 2h, h = a/2

b = 2k, k = b/2

The co-ordinates of the centre areÂ (a/2, b/2)

**4. Find the equation of the circle which touches x-axis and whose centre is (1, 2).**

**Solution:**

Since the circle has a centre (1, 2) and also touches x-axis.

Radius of the circle is, r = 2

The equation of a circle having centre (h, k), having radius as r units, is

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

So, the equation of the required circle is:

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

â‡’x^{2}Â – 2x + 1 + y^{2}Â – 4y + 4 = 4

â‡’Â x^{2}Â + y^{2}Â â€“ 2x – 4y + 1 = 0

The equation of the circle isÂ x^{2}Â + y^{2}Â â€“ 2x – 4y + 1 = 0.

**5. If the lines 3x â€“ 4y + 4 = 0 and 6x â€“ 8y â€“ 7 = 0 are tangents to a circle, then find the radius of the circle.**

**Solution:**

**6. Find the equation of a circle which touches both the axes and the line 3x â€“ 4y + 8 = 0 and lies in the third quadrant.**

**Solution:**

The equation of the given circle is x^{2}Â + y^{2}Â + 4x + 4y + 4 = 0.

**7. If one end of a diameter of the circle x ^{2}Â + y^{2}Â â€“ 4x â€“ 6y + 11 = 0 is (3, 4), then find the coordinate of the other end of the diameter.**

**Solution:**

Given equation of the circle,

x^{2}Â – 4x + y^{2}Â – 6y + 11 = 0

x^{2}Â – 4x + 4 + y^{2}Â – 6y + 9 +11 â€“ 13 = 0

the above equation can be written as

x^{2}Â â€“ 2 (2) x + 2^{2}Â + y^{2}Â â€“ 2 (3) y + 3^{2}Â +11 â€“ 13 = 0

on simplifying we get

(x â€“ 2)^{2}Â + (y â€“ 3)^{2}Â = 2

(x â€“ 2)^{2}Â + (y â€“ 3)^{2}Â =Â (âˆš2)^{2}

Since, the equation of a circle having centre (h, k), having radius as r units, is

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

We have centre = (2, 3)

The centre point is the mid-point of the two ends of the diameter of a circle.

Let the points be (p, q)

(p + 3)/2 = 2 and (q + 4)/2 = 3

p + 3 = 4 & q + 4 = 6

p = 1 & q = 2

Hence, the other ends of the diameter are (1, 2).

**8. Find the equation of the circle having (1, â€“2) as its centre and passing through 3x + y = 14, 2x + 5y = 18**

**Solution:**

Solving the given equations,

3x + y = 14 â€¦â€¦â€¦.1

2x + 5y = 18 â€¦â€¦â€¦â€¦..2

Multiplying the first equation by 5, we get

15x + 5y = 70â€¦â€¦.3

2x + 5y = 18â€¦â€¦..4

Subtract equation 4 from 3 we get

13 x = 52,

Therefore x = 4

Substituting x = 4, in equation 1, we get

3 (4) + y = 14

y = 14 â€“ 12 = 2

So, the point of intersection is (4, 2)

Since, the equation of a circle having centre (h, k), having radius as r units, is

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

Putting the values of (4, 2) and centre co-ordinates (1,-2) in the above expression, we get

(4 â€“ 1)^{2}Â + (2 â€“ (-2))^{2}Â = r^{2}

3^{2}Â + 4^{2}Â = r^{2}

r^{2}Â = 9 + 16 = 25

r = 5 units

So, the expression is

(x â€“ 1)^{2}Â + (y â€“ (-2))^{2}Â = 5^{2}

Expanding the above equation we get

x^{2}Â – 2x + 1 + (y + 2)^{2}Â = 25

x^{2}Â – 2x + 1 + y^{2}Â + 4y + 4 = 25

x^{2}Â – 2x + y^{2}Â + 4y â€“ 20 = 0

Hence the required expression is x^{2}Â – 2x + y^{2}Â + 4y â€“ 20 = 0.

**9. If the line y = âˆš3x + kÂ touches the circle x ^{2}Â + y^{2}Â = 16, then find the value of k.**

**Solution:**

Hence, the required value of k is 8.

**10. Find the equation of a circle concentric with the circle x ^{2}Â + y^{2}Â â€“ 6x + 12y + 15 = 0 and has double of its area.**

**Solution:**

Given equation of the circle is

x^{2}Â – 6x + y^{2}Â + 12y + 15 = 0

The above equation can be written as

x^{2}Â â€“ 2 (3) x + 3^{2}Â + y^{2}Â + 2 (6) y + 6^{2}Â + 15 â€“ 9 + 36 = 0

(x â€“ 3)^{2}Â + (y-(-6))^{2}Â – 30 = 0

(x â€“ 3)^{2}Â + (y-(-6))^{2}Â =Â (âˆš30)^{2}

Since, the equation of a circle having centre (h, k), having radius as r units, is

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

Centre = (3,-6)

Area of inner circle = Ï€r^{2} = 22/7 Ã— 30 = 30 Ï€ units square

Area of outer circle = 2 Ã— 30 Ï€ =Â 60 Ï€ units square

So, Ï€r^{2} = 60 Ï€

r^{2}Â = 60

Equation of outer circle is,

(x â€“ 3)^{2}Â + (y – (-6))^{2}Â =Â (âˆš60)^{2}

x^{2}Â – 6x + 9 + y^{2}Â – 12 y + 36 = 60

x^{2}Â – 6x + y^{2}Â +12y +45 â€“ 60 = 0

x^{2}Â – 6x + y^{2}Â + 12y â€“ 15 = 0

Hence, the required equation of the circle isÂ x^{2}Â – 6x + y^{2}Â + 12y â€“ 15 = 0.

**11. If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity.**

**Solution:**

**12. Given the ellipse with equation 9x ^{2} + 25y^{2} = 225, find the eccentricity and foci.**

**Solution:**

**13. If the eccentricity of an ellipse is 5/8 and the distance between its foci is 10, then find latus rectum of the ellipse.**

Solution:

**14. Find the equation of ellipse whose eccentricity is 2/3 , latus rectum is 5 and the centre is (0, 0).**

**Solution:**

**15. Find the distance between the directrices of the ellipse x ^{2}/36 + y^{2}/20 = 1**

**Solution:**

**16. Find the coordinates of a point on the parabola y2 = 8x whose focal distance is 4.**

**Solution:**

We know that equation of an ellipse is y^{2}Â = 4ax,

Also we have length of latus rectum = 4a

Now by comparing the above two equations,

4a = 8

Therefore

a = 2

y^{2 }= 8 Ã— 2 = 16

y = Â± 4 and x = 2

Hence the co – ordinates are (2, 4) and (2, -4).

**17. Find the length of the line-segment joining the vertex of the parabola y2 = 4axand a point on the parabola where the line-segment makes an angle q to the x-axis.**

**Solution:**

**18. If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.**

**Solution:**

Given

**19. If the line y = mx + 1 is tangent to the parabola y2 = 4x then find the value of m.**

**Solution:**

Given equations are,

y = mx + 1 & y^{2}Â = 4x

By solving given equations we get

(mx + 1)^{2}Â = 4x

Expanding the above equation we get

m^{2}x^{2}Â + 2mx + 1 = 4x

On rearranging we get

m^{2}x^{2}Â + 2mx â€“ 4x + 1 = 0

m+x^{2}Â + x (2m â€“ 4) + 1 = 0

As the line touches the parabola, above equation must have equal roots,

Discriminant (D) = 0

(2m â€“ 4)^{2}Â – 4 (m^{2}) (1) = 0

4m^{2}Â – 16m + 16 â€“ 4m^{2}Â = 0

-16 m + 16 = 0

– m + 1 = 0

m = 1

Hence, the required value of m is 1.

**20. If the distance between the foci of a hyperbola is 16 and its eccentricity is âˆš2, then obtain the equation of the hyperbola.**

**Solution:**

**21. Find the eccentricity of the hyperbola 9y ^{2}Â â€“ 4x^{2}Â = 36.**

**Solution:**

**22. Find the equation of the hyperbola with eccentricity 3/2 and foci at (Â± 2, 0).**

**Solution:**

Long Answer type:

**23. If the lines 2x â€“ 3y = 5 and 3x â€“ 4y = 7 are the diameters of a circle of area 154square units, then obtain the equation of the circle.**

**Solution:**

Since, diameters of a circle intersect at the centre of a circle,

2x â€“ 3y = 5 â€¦â€¦â€¦1

3x â€“ 4y = 7 â€¦â€¦â€¦..2

Solving the above equations,

Multiplying equation 1 by 3 we get

6x â€“ 9y = 15

Multiplying equation 2 by 2 we get

6x â€“ 8y = 14

y = 1

y = -1

Putting y = -1, in equation 1, we get

2x â€“ 3(-1) = 5

2x + 3 = 5

2x = 2

x = 1

Coordinates of centre = (1,-1)

Given area = 154

Area = Ï€r^{2}Â = 154

22/7 Ã— r^{2} = 154

r^{2} = 154 Ã— 7/22

r = 7 units

Since, the equation of a circle having centre (h, k), having radius as r units, is

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

(x â€“ 1)^{2}Â + (y â€“ (-1))^{2}Â = 7^{2}

x^{2}Â – 2x +1 + (y + 1)^{2}Â = 49

x^{2}Â – 2x + 1 + y^{2}Â + 2y + 1 â€“ 49 = 0

x^{2}Â – 2x + y^{2}Â + 2y â€“ 47 = 0

Hence the required equation of the given circle is x^{2}Â – 2x + y^{2}Â + 2y â€“ 47 = 0.

**24. Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y â€“ 4x + 3 = 0.**

**Solution:**

Since, the equation of a circle having centre (h, k), having radius as r units, is

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

Substituting (2, 3) & (4, 5) in the above equation, we get

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

4 â€“ 4h + h^{2}Â + 9 + k^{2}Â – 6k = r^{2}

h^{2}Â – 4h + k^{2}Â – 6k + 13 = r^{2}Â â€¦â€¦â€¦â€¦..2

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

16 â€“ 8h + h^{2}Â + 25 + k^{2}Â – 10k = r^{2}

h^{2}Â – 8h + k^{2}Â – 10k + 41 = r^{2}Â â€¦â€¦â€¦..3

Equating both the equations 2 & 3, as their RHS are equal, we get

h^{2}Â – 4h + k^{2}Â – 6k + 13 = h^{2}Â – 8h + k^{2}Â – 10k + 41

On simplifying we get

8h – 4h + 10k â€“ 6k = 41 â€“ 13

4h + 4k = 28

h + k = 7 â€¦â€¦â€¦..4

As centre lies on the given line, so it satisfies the values too,

k â€“ 4h + 3 = 0â€¦â€¦â€¦â€¦5

Solving equations 3 and 4 simultaneously,

h + k = 7

-4h + k = -3

Subtracting both the equations, we get

5h = 10

h = 2

2 + k = 7

k = 5

Putting h = 2 & k = 5 in equation 2,

h^{2}Â – 4h + k^{2}Â – 6k + 13 = r^{2}

2^{2}Â – 4(2) + 5^{2}Â – 6(5) + 13 = r^{2}

4 â€“ 8 + 25 â€“ 30 + 13 = r^{2}

r^{2}Â = 4

r = 2 units

Putting the values of h = 2, k = 5 & r = 2, respectively in equation 1,

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

(x â€“ 2)^{2}Â + (y â€“ 5)^{2}Â = 2^{2}

x^{2}Â – 4x + 4 + y^{2}Â – 10y + 25 = 4

x^{2}Â – 4x + y^{2}Â – 10y + 25 =0

Hence, the required equation is x^{2}Â – 4x + y^{2}Â – 10y + 25 = 0.

**25. Find the equation of a circle whose centre is (3, â€“1) and which cuts off a chord of length 6 units on the line 2x â€“ 5y + 18 = 0.**

**Solution:**

Using Pythagoras Theorem,

(Hypotenuse)^{2}Â = (Base)^{2}Â + (Perpendicular)^{2}

= (3)^{2}Â + (âˆš29)^{2}Â = 29 + 9

= âˆš38

Hypotenuse = âˆš38 units (radius)

Since, the radius bisects the chord into two equal halves,

Since, the equation of a circle having centre (h, k), having radius as r units, is

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

(x â€“ 3)^{2}Â + (y â€“ (-1))^{2}Â = (âˆš38)^{2}

x^{2}Â – 6x + 9 + (y + 1)^{2}Â = 38

x^{2}Â – 6x + y^{2}Â + 2y + 1 + 9- 38 = 0

x^{2}Â – 6x + y^{2}Â + 2y â€“ 28 = 0

Hence, the required equation of the circle is x^{2}Â – 6x + y^{2}Â + 2y â€“ 28 = 0.

**26. Find the equation of a circle of radius 5 which is touching another circle x ^{2} + y^{2} â€“ 2x â€“ 4y â€“ 20 = 0 at (5, 5).**

**Solution:**

Given x^{2}Â – 2x + y^{2}Â – 4y â€“ 20 = 0

x^{2}Â – 2x + 1 +y^{2}Â – 4y + 4 â€“ 20 â€“ 5 = 0

(x â€“ 1)^{2}Â + (y â€“ 2)^{2}Â = 25

(x â€“ 1)^{2}Â + (y â€“ 2)^{2}Â = 5^{2}

Since, the equation of a circle having centre (h, k), having radius as r units, is

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

Centre = (1, 2)

Point of Intersection = (5, 5)

It intersects the line into 1: 1, as the radius of both the circles is 5 units.

Using Ratio Formula,

p + 1 = 10, q + 2 = 10

p = 9 & q = 8

Co-ordinates = (9, 8)

Therefore the equation is,

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

(x â€“ 9)^{2}Â + (y â€“ 8)^{2}Â = 5^{2}

x^{2}Â – 18x + 81 + y^{2}Â – 16y + 64 = 25

x^{2}Â – 18x + y^{2}Â – 16y + 145 â€“ 25 = 0

x^{2}Â – 18x + y^{2}Â – 16y + 120 = 0

Hence, the required equation is x^{2}Â – 18x + y^{2}Â – 16y + 120 = 0.

**27. Find the equation of a circle passing through the point (7, 3) having radius 3 units and whose centre lies on the line y = x â€“ 1.**

**Solution:**

Since, the equation of a circle having centre (h, k), having radius as r units, is

(x â€“ h)^{2}Â + (y â€“ k)^{2}Â = r^{2}Centre lies on the line i.e., y = x â€“ 1,

Co â€“ Ordinates are (h, k) = (h, h â€“ 1)

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

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

49 + h^{2}Â – 14h + (3 â€“ h +1)^{2}Â = 9

On rearranging we get

h^{2}Â – 14h + 49 +16 +h^{2}Â – 8h â€“ 9 = 0

2h^{2}Â – 22h + 56 = 0

h^{2}Â – 11h + 28 = 0

h^{2}Â – 4h â€“ 7h + 28 = 0

h (h â€“ 4) â€“ 7 (h â€“ 4) = 0

(h â€“ 7) (h â€“ 4) = 0

h = 7 or 4

Centre = (7, 6) or (4, 3)

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

Equation, having centre (7, 6)

(x â€“ 7)^{2}Â + (y â€“ 6)^{2}Â = 3^{2}

x^{2}Â – 14x + 49 + y^{2}Â – 12y + 36 â€“ 9 = 0

x^{2}Â – 14x + y^{2}Â – 12y + 76 = 0

Equation, having centre (4, 3)

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

x^{2}Â – 8x + 16 + y^{2}Â – 6y + 9 â€“ 9 = 0

x^{2}Â – 8x + y^{2}Â – 6y + 16 = 0

Hence, the required equation isÂ x^{2}Â – 14x + y^{2}Â – 12y + 76 = 0 or x^{2}Â – 8x + y^{2}Â – 6y + 16 = 0.

**28. Find the equation of each of the following parabolas **

**(a) Directrix x = 0, focus at (6, 0) **

**(b) Vertex at (0, 4), focus at (0, 2) **

**(c) Focus at (â€“1, â€“2), directrix x â€“ 2y + 3 = 0**

**Solution:**

(a) The distance of any point on the parabola from its focus and its directrix is same.

Given that, directrix, x = 0 and focus = (6, 0)

If a parabola has a vertical axis, the standard form of the equation of the parabola is (x – h)^{2}Â = 4p (y – k), where pâ‰ 0.

The vertex of this parabola is at (h, k).

The focus is at (h, k + p) & the directrix is the line y = k – p.

As the focus lies on x â€“ axis,

Therefore the equation is y^{2}Â = 4ax or y^{2}Â = -4ax

So, for any point P(x, y) on the parabola

Distance of point from directrix = Distance of point from focus

x^{2}Â = (x â€“ 6)^{2}Â + y^{2}

x^{2}Â = x^{2}Â – 12x + 36 + y^{2}

y^{2}Â – 12x + 36 = 0

Hence the required equation is y^{2}Â – 12x + 36 = 0.