**Important questions for class 11 Maths Chapter 11** – conic sections with solutions are given here. The important questions given here are based on the latest exam pattern and previous year question papers and sample papers. Solving these questions will help the students to score good marks in the annual examinations. The case study questions are framed as per the **CBSE board syllabus (2022-2023)**Â and **NCERT curriculum**.Â Also, HOTS and value-based questions are asked related to the concept.

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Class 11 Maths Chapter 11 – conic sections will incorporate the concept of sections of cone such as

- Parabola
- Ellipse
- Hyperbola
- Circle

Learn about these conic sections by solving the questions here. At the end of the article, we have also provided extra questions for chapter 11 conic sections, so that students can practice more on the topic.

## Class 11 Chapter 11 – Conic Sections Important Questions with Solutions

To score the good marks in the final examination, practice the problems provided here, which will help you to solve the problems in the annual examination.

**Question 1:**

Determine the equation of the circle with radius 4 and Centre (-2, 3).

**Solution:**

Given that:

Radius, r = 4, and center (h, k) = (-2, 3).

We know that 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} â€¦.(1)

Now, substitute the radius and center values in (1), we get

Therefore, the equation of the circle is

(x + 2)^{2}+ (y – 3)^{2} = (4)^{2}

x^{2}+ 4x + 4 + y^{2} – 6y + 9 = 16

Now, simplify the above equation, we get:

x^{2} + y^{2}+ 4x – 6y – 3 = 0

Thus, the equation of a circle with center (-2, 3) and radius 4 is x^{2} + y^{2}+ 4x – 6y – 3 = 0

**Question 2: **

Compute the centre and radius of the circle 2x^{2} + 2y^{2} – x = 0

**Solution: **

Given that, the circle equation is 2x^{2} + 2y^{2} – x = 0

This can be written as:

â‡’ (2x^{2}-x) + y^{2} = 0

â‡’ 2{[x^{2} – (x/2)] +y^{2}} = 0

â‡’{ x^{2} – 2x(Â¼) + (Â¼)^{2}} +y^{2} – (Â¼)^{2} = 0

Now, simplify the above form, we get

â‡’(x- (Â¼))^{2} + (y-0)^{2} = (Â¼)^{2}

The above equation is of the form (x â€“ h)^{2} + (y â€“ k)^{2}= r^{2}

Therefore, by comparing the general form and the equation obtained, we can say

h= Â¼ , k = 0, and r = Â¼.

**Question 3:**

Determine the focus coordinates, the axis of the parabola, the equation of the directrix and the latus rectum length for y^{2} = -8x

**Solution:**

Given that, the parabola equation is y^{2}= -8x.

It is noted that the coefficient of x is negative.

Therefore, the parabola opens towards the left.

Now, compare the equation with y^{2}= -4ax, we obtain

-4a= -8

â‡’ a = 2

Thus, the value of a is 2.

Therefore, the coordinates of the focus = (-a, 0) = (-2, 0)

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

Equation of directrix, x= a i.e., x = 2

We know the formula to find the length of a latus rectum

Latus rectum length= 4a

Now, substitute a = 2, we get

Length of latus rectum = 8

**Question 4: **

Determine the foci coordinates, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse (x^{2}/49) + (y^{2}/36) = 1

**Solution:**

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

It can be written as (x^{2}/7^{2}) + (y^{2}/6^{2}) = 1

It is noticed that the denominator of x^{2}/49 is greater than the denominator of the y^{2}/36

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

a= 7 and b = 6

Therefore, c = âˆš(a^{2}– b^{2})

Now, substitute the value of a and b

â‡’ âˆš(a^{2}– b^{2}) = âˆš(7^{2}– 6^{2}) = âˆš(49-36)

â‡’ âˆš13

Hence, the foci coordinates are ( Â± âˆš13, 0)

Eccentricity, e = c/a = âˆš13/ 7

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

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

The coordinates of the vertices are ( Â± 7, 0)

Latus rectum Length= 2b^{2}/a = 2(6)^{2}/7 = 2(36)/7 = 72/7

**Question 5:**

Determine the equation for the ellipse that satisfies the given conditions: Centre at (0, 0), the major axis on the y-axis and passes through the points (3, 2) and (1, 6).

**Solution:**

Centre = (0, 0), and major axis that passes through the points (3, 2) and (1, 6).

We know that the equation of the ellipse will be of the form when the centre is at (0, 0) and the major axis is on the y-axis,

(x^{2}/b^{2}) + (y^{2}/a^{2}) = 1 â€¦. (1)

Here, a is the semi-major axis.

It is given that, the ellipse passes through the points (3, 2) and (1, 6).

Hence, equation (1) becomes

(9/b^{2}) + (4/a^{2}) = 1 â€¦(2)

(1/b^{2}) + (36/a^{2}) = 1 â€¦(3)

Solving equation (2) and (3), we get

b^{2} = 10 and a^{2} = 40

Therefore, the equation of the ellipse becomes: (x^{2}/10) + (y^{2}/40) = 1

**Question 6:**

Determine the equation of the hyperbola which satisfies the given conditions: Foci (0, Â±13), the conjugate axis is of length 24.

**Solution:**

Given that: Foci (0, Â±13), Conjugate axis length = 24

It is noted that the foci are on the y-axis.

Therefore, the equation of the hyperbola is of the form:

(y^{2}/a^{2})-(x^{2}/b^{2}) = 1 â€¦(1)

Since the foci are (0, Â±13), we can get

C = 13

It is given that, the length of the conjugate axis is 24,

It becomes 2b = 24

b= 24/2

b= 12

And, we know that a^{2} + b^{2} = c^{2}

To find a, substitute the value of b and c in the above equation:

a^{2} + 12^{2} = 13^{2}

a^{2 }= 169-144

a^{2} = 25

Now, substitute the value of a and b in equation (1), we get

(y^{2}/25)-(x^{2}/144) = 1, which is the required equation of the hyperbola.

## More Articles for Class 11

- Class 11 Syllabus
- Important 1 Mark Questions for CBSE Class 11 Maths
- Important 4 Marks Questions for CBSE Class 11 Maths
- Important 6 Marks Questions for CBSE Class 11 Maths
- Tips to score better marks in class 11 Maths Exam

## Practice Problems for Class 11 Maths Chapter 11 Conic Sections

These class 11 Conic Sections questions are categorized into short answer type questions and long answer type questions. These extra questions cover various concepts which will help class 11 students to develop problem-solving skills for the exam.

- Calculate the equation of a circle that passes through the origin and cuts off intercepts -2 and 3 from the axis and the y-axis respectively. (Solution: x
^{2}+ y^{2}+ 2x -3y) - Determine the equation of the circle passing through the points – (0,0)(5,0) and (3,3). (Solution: x
^{2}+ y^{2}– 5x -y =0), centre (5/2 , Â½) and radius = âˆš 26/2). - If the distance between the foci of a hyperbola is 16 and eccentricity is âˆš 2, then obtain its equation. (Solution: x
^{2}– y^{2}=32) - If a latus rectum of an ellipse subtends a right angle at the centre of the ellipse, then write the eccentricity of the ellipse. (Solution: (âˆš 5 – 1) / 2)
- Determine the equation of the ellipse whose foci are (4,0) and (-4,0), eccentricity = â…“. (Solution: x
^{2}/ 9 + y^{2}/8 = 16) - Write the equation of the parabola whose vertex is at (-3,0) and the directrix is (x + 5 ) = 0. (Solution: y
^{2}= 8(x + 3)) - AB is a double ordinate of a parabola y
^{2}= 4px. Find the locus of its points of trisection. (Solution: 9y^{2}=4px) - Calculate the equation of the parabola whose focus is (1, -1) and whose vertex is (2,1). Also, find its axis and latus- rectum). (Solution: 4 âˆš 5).
- Find the equation of the circle which circumscribes the triangle formed by the lines x = 0, y = 0 and lx +my = 1. (Solution: x
^{2}+ y^{2}– (1/l)x – (1/m)y = 0) - Prove that the points (9,1) ( 7,9) (-2, 12) and (6,10) are concyclic.
- Find the equation of ellipse whose eccentricity is 2/3, latus rectum is 5 and the centre is (0,0).
- Find the equation of the circle which touches x-axis and whose centre is (1,2).
- Find the coordinates of a point on the parabola y
^{2}=8x whose focal distance is 4.

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