NCERT Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.1, Ex 11.2, Ex 11.3, Ex 11.4, Ex 11.5, Miscellaneous Exercise
NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.1
Ex 11.1 Class 11 Question 1:![]()
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Ex 11.1 Class 11 Maths Question 11:
Find the equation of the circle passing through the points (2, 3) and (-1, 1) and whose centre is on the line x – 3y – 11 = 0.
Solution:
The equation of the circle is,
(x – h)2 + (y – k)2 = r2 ….(i)
Since the circle passes through point (2, 3)
∴ (2 – h)2 + (3 – k)2 = r2
⇒ 4 + h2 – 4h + 9 + k2 – 6k = r2
⇒ h2+ k2 – 4h – 6k + 13 = r2 ….(ii)
Also, the circle passes through point (-1, 1)
∴ (-1 – h)2 + (1 – k)2 = r2
⇒ 1 + h2 + 2h + 1 + k2 – 2k = r2
⇒ h2 + k2 + 2h – 2k + 2 = r2 ….(iii)
From (ii) and (iii), we have
h2 + k2 – 4h – 6k + 13 = h2 + k2 + 2h – 2k + 2
⇒ -6h – 4k = -11 ⇒ 6h + 4k = 11 …(iv)
Since the centre (h, k) of the circle lies on the line x – 3y-11 = 0.
∴ h – 3k – 11 = 0 ⇒ h -3k = 11 …(v)
Solving (iv) and (v), we get
h = 72 and k = −52
Putting these values of h and k in (ii), we get
(72)2+(−52)2−4×72−6×−52+13=r2
⇒ 494+254−14+15+13 ⇒ r2=652
Thus required equation of circle is
⇒ (x−72)2+(y+52)2=652
⇒ x2+494−7x+y2+254+5y=652
⇒ 4×2 + 49 – 28x + 4y2 + 25 + 20y = 130
⇒ 4×2 + 4y2 – 28x + 20y – 56 = 0
⇒ 4(x2 + y2 – 7x + 5y -14) = 0
⇒ x2 + y2 – 7x + 5y -14 = 0.
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NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.2
Ex 11.2 Class 11 Maths Question 1:![]()
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NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.3
Ex 11.3 Class 11 Maths Question 1:
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NCERT Solutions for Class 11 Maths Chapter 11 Exercise 11.4
Ex 11.4 Class 11 Maths Question 1:
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Ex 11.4 Class 11 Maths Question 8.
Vertices (0, ±5), foci (0, ±8)
Solution:
Vertices are (0, ±5) which lie on x-axis. So the equation of hyperbola in standard form
Ex 11.4 Class 11 Maths Question 9.
Vertices (0, ±3), foci (0, ±5)
Solution:
Vertices are (0, ±3) which lie on x-axis. So the equation of hyperbola in standard form

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Class 11 Maths NCERT Solutions – Miscellaneous Questions
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