# Important Questions for Class 11 Chapter 11 - Conic Sections

Students should practice a lot of questions based on their NCERT syllabus to get a hold of the concepts learned. This Important Questions for class 11 strictly adheres to the guidelines shared by CBSE. Thus, you can benefit a lot by practicing the questions. The more you practice, the better marks you get in the examinations.

Important questions based on NCERT syllabus for Class 11 Chapter 11 - Conic Sections:

Question 1: Show that the equation x^2 + y^2 + 8x -5y - 49 = 0 represets the circle and also find the centre and radius.

Solution: We know that the equation for circle is x^2 + y^2 + 2gx -2fy - c = 0

Given x^2 + y^2 + 8x -5y - 49 = 0

Here 2g = 8 => g = 8/2 = 4
And 2f = -5 => f = -5/2 and c = -49

So centre of the circle = (-g, -f) = (-4, -(-5/2)) = (-4,5/2)

And Radius of the circle = √ g^2 +f^2 - c
= √ 4^2 + (5/2)^2 - 49
= √ 16 + 25/4 - 49
= √ - 33 + 25/4
= √ (-132+ 25)/4
= √ (-107)/4
= 1/2 √(-107)

Question 2: Find the equation of the ellipse whose vertices are (0, ±9) and e = 5/6

Solution: Let the equation of the ellipse be x^2/b^2 + y^2/a^2 = 1 and its vertice are (0, ±a)

here a = ±9

and let c^2 = a^2 - b^2 then e = c/a
=> c = ea
=> c = 9x5/6
=> c = 15/2
Now c^2 = a^2 - b^2
=> b^2 = a^2 - c^2
=> b^2 = 9^2 - (15/2)^2
=> b^2 = 81 - 225/4
=> b^2 = (324 - 225)/4
=> b^2 = 99/4
=> b = 1/2√99

Substitute the values of a and b in the ellipse equation we get

2x^2/√99 + y^2/91 = 1

Hence the equation of the ellipse is 2x^2/√99 + y^2/91 = 1

Question 3: Find the equation of the circle with radis 6cm whose centre lie on x axis and passing through the points (4, 5)

Solution:

Let the equation of the circle be (x-h)^2 + (y-k)^2 = r^2

=> (4-h)^2 + (5-k)^2 = 6^2
=> 4^2 +h^2 -8h + 5^2 + k^2 - 10k = 36
=> 16 + h^2 - 8h + 25 + k^2 - 10k - 36 = 0
=> h^2 + k^2 - 8h - 10k + 5 = 0..........(1)

Since the centre lie in the x axis we have k = 0

Substitute it in equation 1 we get

h^2 - 8h + 5 = 0
Apply quadratic formula to solve the equation

h = (-b ± √b^2 - 4ac)/2a
h = -(-8) ± √64 - 4(1)(5) / 2
h = 8 ± √ 64 - 20)/2
h = 8 ± √44)/ 2
h = 8 ± √4x11)/2
h = 8 ± 2√11)/2
h = 4 ± √11

Hence the required equation for the circle is

x-4 + √11)^2 + (y-0)^2 = 6^2
x + 4 + √11)^2 + y^2 = 36
x^2 + (4 + √11)^2 + 2(4 + √11)x + y^2 - 36 = 0
x^2 + y^2 + 2(4 + √11)x + (4 + √11)^2 - 36 = 0