NCERT Solutions Class 11 Maths Chapter-11 (Conic Sections)

NCERT Solutions Class 11 Maths from class 11th Students will et the answers of Chapter-11 (Conic Sections) . This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.

NCERT Question-Answe

Class 11 Mathematics

Chapter-11 (Conic Sections)

Chapter-11 (Conic Sections)

Exercise 11.1

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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.
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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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Exercise 11.2

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Exercise 11.3

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Exercise 11.4

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Question 8.
Vertices (0, ±5), foci (0, ±8)
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Vertices are (0, ±5) which lie on x-axis. So the equation of hyperbola in standard form

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Vertices (0, ±3), foci (0, ±5)
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Vertices are (0, ±3) which lie on x-axis. So the equation of hyperbola in standard form

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