# NCERT Solutions for Class 11 Maths Chapter 11

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 Class: 11 Subject: Maths Chapter 11: Conic Sections

## NCERT Solutions for Class 11 Maths Chapter 11

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### Class 11 Maths Chapter 11 Conic Sections Solutions

#### Important Terms on Conic Sections

• Circle, ellipse, parabola and hyperbola are curves which are obtained by intersection of a plane and cone in different positions
• Circle: It is the set of all points in a plane that are equidistant from a fixed point in that plane.
• Parabola: It is the set of all points in a plane which are equidistant from a fixed point (focus) and a fixed line in the plane. Fixed point does not lie on the line.
• Latus Rectum: A chord through focus perpendicular to axis of parabola is called its latus rectum.
• Ellipse: It is the set of points in a plane the sum of whose distances from two fixed points in the plane is a constant and is always greater than the distances between the fixed points.
• Hyperbola: It is the set of all points in a plane, the differences of whose distance from two fixed points in the plane is a constant.

##### Important Extra Questions with Answers on Conic Sections
1. Find the centre and radius of the circle x² + y² – 6x + 4y – 12 = 0. [Answer: (3, -2), 5]
2. Find the equation of hyperbola satisfying given conditions foci (5, 0) and transverse axis is of length 8. [Answer: x²/16 – y²/9 = 1.]
3. Find the coordinates of points on parabola y² = 8x whose focal distance is 4. [Answer: (2, 4), (2, -4)]
4. If one end of a diameter of the circle x² + y² – 4x – 6y + 11 = 0 is (3, 4), then find the coordinates of the other end of diameter. [Answer: (1, 2)]
5. Find equation of an ellipse having vertices (0, 5) and foci (0, 4). [Answer: x²/9 + y²/25 = 1]
6. Find the equation of a circle whose centre is at (4, –2) and 3x – 4y+ 5 = 0 is tangent to circle. [Answer: x² + y² – 8x + 4y – 5 = 0]
7. If the distance between the foci of a hyperbola is 16 and its eccentricity is 2, then obtain the equation of a hyperbola. [Answer: x² – y² = 32]
8. If the latus rectum of an ellipse is equal to half of minor axis, then find its eccentricity. [Answer: e = √3/2]

###### Try These
• Find the length of major and minor axis of the following ellipse, 16x² + 25y² = 400. [Answer: 10, 8]
• Find equation of circle concentric with circle 4x² + 4y² – 12x – 16y – 21 = 0 and of half its area. [Answer: 2x² + 2y² – 6x + 8y + 1 = 0]
• Find the equation for the ellipse that satisfies the given condition Major axis on the x-axis and passes through the points (4, 3) and (6, 2). [Answer: x²/52 + y²/13 = 1]

#### Find the equation of the circle with centre (0, 2) and radius 2.

The equation of a circle with centre (h, k) and radius r is given as
(x – h)2 + (y – k)2 = r2
It is given that centre (h, k) = (0, 2) and radius (r) = 2.
Therefore, the equation of the circle is
(x – 0)2 + (y – 2)2 = 22
⟹ x2 + y2 + 4 – 4 y = 4
⟹ x2 + y2 – 4y = 0

#### Find the equation of the circle with centre (–2, 3) and radius 4.

The equation of a circle with centre (h, k) and radius r is given as
(x – h)^2 + (y – k)^2 = r^2
It is given that centre (h, k) = (–2, 3) and radius (r) = 4.
Therefore, the equation of the circle is (x + 2)2 + (y – 3)2 = (4)2
⟹ x^2 + 4x + 4 + y^2 – 6y + 9 = 16
⟹ x^2 + y^2 + 4x – 6y – 3 = 0

#### Find the centre and radius of the circle (x + 5)^2 + (y – 3)^2 = 36

The equation of the given circle is (x + 5)^2 + (y – 3)^2 = 36.
(x + 5)^2 + (y – 3)^2 = 36
⇒ {x – (–5)}^2 + (y – 3)^2 = 62,
which is of the form (x – h)^2 + (y – k)^2 = r^2,
where h = – 5, k = 3, and r = 6.
Thus, the centre of the given circle is (–5, 3), while its radius is 6.

#### Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.

Let the equation of the required circle be (x – h)^2 + (y – k)^2 = r^2.
Since the circle passes through points (4, 1) and (6, 5),
(4 – h)^2 + (1 – k)^2 = r^2 …… (1)
(6 – h)^2 + (5 – k)^2 = r^2 …… (2)
Since the centre (h, k) of the circle lies on line 4x + y = 16,
4h + k = 16 …… (3)
From equations (1) and (2), we obtain
(4 – h)^2 + (1 – k)^2 = (6 – h)^2 + (5 – k)^2
⇒ 16 – 8h + h^2 + 1 – 2k + k^2 = 36 – 12h + h^2 + 25 – 10k + k^2
⇒ 16 – 8h + 1 – 2k = 36 – 12h + 25 – 10k
⇒ 4h + 8k = 44
⇒ h + 2k = 11 …… (4)
On solving equations (3) and (4), we obtain h = 3 and k = 4.
On substituting the values of h and k in equation (1), we obtain
(4 – 3)^2 + (1 – 4)^2 = r^2
⇒ (1)^2 + (– 3)^2 = r^2
⇒ 1 + 9 = r^2
⇒ r^2 = 10
⇒ 𝑟=√10
Thus, the equation of the required circle is
(x – 3)^2 + (y – 4)^2 = (√10)^2
x^2 – 6x + 9 + y^2 – 8y + 16 = 10
x^2 + y^2 – 6x – 8y + 15 = 0

#### Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y^2 = 12x.

The given equation is y^2 = 12x.
Here, the coefficient of x is positive. Hence, the parabola opens towards the right.
On comparing this equation with y^2 = 4ax, we obtain
4a = 12
⇒ a = 3
∴ Coordinates of the focus = (a, 0) = (3, 0)
Since the given equation involves y^2, the axis of the parabola is the x-axis.
Equation of direcctrix,
x = –a i.e., x = – 3 i.e., x + 3 = 0
Length of latus rectum = 4a = 4 × 3 = 12.

#### Find the equation of the parabola that satisfies the following conditions: Focus (6, 0); directrix x = –6.

Focus (6, 0); directrix, x = –6
Since the focus lies on the x-axis, the x-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form y2 = 4ax or
y^2 = – 4ax.
It is also seen that the directrix, x = –6 is to the left of the y-axis, while the focus (6, 0) is to the right of the y-axis.
Hence, the parabola is of the form y^2 = 4ax.
Here, a = 6
Thus, the equation of the parabola is y^2 = 24x.