Inter 1st Year Maths Conic Sections Solutions Exercise 10d

Practicing the AP Board Solutions Class 11 Maths and Chapter 10 Inter 1st Year Maths Conic Sections Solutions Exercise 10d Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Conic Sections Solutions Exercise 10d

Conic Sections Exercise 10d Solutions

Conic Sections Class 11 Exercise 10d Solutions – Conic Sections 10d Exercise Solutions

I.

Question 1.
Find the eccentricity of hyperbola \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\)
Solution:
For the hyperbola \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\)
a² = b² (e² – 1)

Question 2.
Find the eccentricity of a rectangular hyperbola.
Solution:
In an equilateral hyperbola, we have
Length of transverse axis = Length of conjugate axis
⇒ 2a = 2b
⇒ a = b
b² = a² (e² -1)
⇒ a² = a² (e² – 1)
⇒ 1 = e² – 1
⇒ e² = 2
⇒ e = √2

Question 3.
Find the length of the conjugate axis of the hyperbola 3x² – 4y² = 12.
Solution:
Given hyperbola is 3x² – 4y² = 12
⇒ \(\frac{x^2}{4}-\frac{y^2}{3}=1\)
Length of conjugate axis = 2b = 2√3.

Question 4.
Find the length of the transverse axis of the hyperbola x² – 4y² = 4
Solution:
Given hyperbola is x² – 4y² = 4
⇒ \(\frac{x^2}{4}-\frac{y^2}{1}=1\)
Length of transverse axis = 2a
= 2 × 2
= 4

II. In each of the Exercises 1 to 6, find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbolas.

Question 1.
\(\frac{x^2}{16}-\frac{y^2}{9}=1\)
Solution:
\(\frac{x^2}{16}-\frac{y^2}{9}=1\)
Standard Form: \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) (Horizontal transverse axis)
a² = 16 ⇒ a = 4
b² = 9 ⇒ b = 3
∴ c² = a² + b²
⇒ c² = 16 + 9
⇒ c² = 25
⇒ c = 5
Foci: (±c, 0) = (±5, 0)
Vertices: (±a, 0) = (±4, 0)
Length of Transverse Axis (major axis): 2a = 8
Length of Conjugate Axis (minor Axis): 2b = 6
Eccentricity: e = \(\frac{c}{a}=\frac{5}{4}\) = 1.25
Length of Latus Rectum = \(\frac{2 b^2}{a}\) = 2 × \(\frac {9}{4}\)

Question 2.
\(\frac{y^2}{9}-\frac{x^2}{27}=1\)
Solution:
\(\frac{y^2}{9}-\frac{x^2}{27}=1\)
Standard Form: \(\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\) (Vertical transverse axis)
a² = 9 ⇒ a = 3
b² = 27 ⇒ b = 3√3
∴ c² = a² + b²
= 9 + 27
= 36
⇒ c = 6
Foci: (0, ±c) = (0, ±6)
Vertices: (0, ±a) = (0, ±3)
Length of Transverse Axis (major axis): 2a = 6
Length of Conjugate Axis (minor axis): 2b = 2(3√3) = 6√3
Eccentricity: e = \(\frac{c}{a}=\frac{6}{3}\) = 2
Length of Latus Rectum: \(\frac{2 \mathrm{~b}^2}{\mathrm{a}}=\frac{2 \times 27}{3}\) = 18

Question 3.
9y² – 4x² = 36
Solution:
The given equation is 9y² – 4x² = 36
⇒ \(\frac{y^2}{4}-\frac{x^2}{9}=1\)
⇒ \(\frac{y^2}{2^2}-\frac{x^2}{3^2}=1\) ……(1)
On comparing equation (1) with the standard equation of a hyperbola \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\),
we obtain b = 2 and a = 3
We know that a² + b² = c²
⇒ c² = 4 + 9 = 13
⇒ c = √13
The coordinates of the foci are (0, ±√13), and the vertices are (0, ±2).
Eccentricity, e = \(\frac{c}{b}=\frac{\sqrt{13}}{2}\)
Length of latus rectum = \(\frac{2 a^2}{b}=\frac{2 \times 9}{2}\) = 9

Question 4.
16x² – 9y² = 576
Solution:
The given equation is 16x² – 9y² = 576
⇒ \(\frac{x^2}{36}-\frac{y^2}{64}=1\)
⇒ \(\frac{x^2}{6^2}-\frac{y^2}{8^2}=1\) ………(1)
On comparing equation (1) with the standard equation of a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\),
we obtain a = 6 and b = 8.
We know that a² + b² = c²
⇒ c² = 36 + 64 = 100
⇒ c = 10
The coordinates of the foci are (±10, 0), and the vertices are (±6, 0).
Eccentricity, e = \(\frac{c}{a}=\frac{10}{6}=\frac{5}{3}\)
length of latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 64}{6}=\frac{64}{3}\)

Question 5.
5y² – 9x² = 36
Solution:
The given equation is 5y² – 9x² = 36
⇒ \(\frac{y^2}{\left(\frac{36}{5}\right)}-\frac{x^2}{4}=1\)
⇒ \(\frac{y^2}{\left(\frac{6}{\sqrt{5}}\right)^2}-\frac{x^2}{2^2}=1\) ……(1)
On comparing equation (1) with the standard equation of a hyperbola \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\),
we obtain b = \(\frac{6}{\sqrt{5}}\) and a = 2.
We Hnow that a² + b² = c²

Length of latus rectum = \(\frac{2 \mathrm{a}^2}{\mathrm{~b}}=\frac{2 \times 4}{\left(\frac{6}{\sqrt{5}}\right)}=\frac{4 \sqrt{5}}{3}\)

Question 6.
49y² – 16x² = 784
Solution:
The given equation is 49y² – 16x² = 784
⇒ \(\frac{y^2}{16}-\frac{x^2}{49}=1\)
⇒ \(\frac{y^2}{4^2}-\frac{x^2}{7^2}=1\) ……..(1)
On comparing equation (1) with the standard equation of a hyperbola \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\)
we obtain b = 4 and a = 7.
We know that a² + b² = c²
⇒ c² = 16 + 49 = 65
⇒ c = √65
The coordinates of the foci are (0, ±√65), and the coordinates of the vertices are (0, ±4).
Eccentricity, e = \(\frac{c}{b}=\frac{\sqrt{65}}{4}\)
Length of latus rectum = \(\frac{2 \mathrm{a}^2}{\mathrm{~b}}=\frac{2 \times 49}{4}=\frac{49}{2}\)

In each of the Exercises 7 to 16, find the equations of the hyperbola satisfying the given conditions.

Question 7.
Vertices (±2, 0), foci (±3, 0)
Solution:
Vertices (±2, 0), foci (±3, 0)
Here, the vertices are on the x-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\).
Since the vertices are (±2, 0)
⇒ a = 2
Since the foci are (±3, 0)
⇒ c = 3
We know that a² + b² = c²
⇒ 2² + b² = 3²
⇒ b² = 9 – 4
⇒ b² = 5
Thus, the equation of the hyperbola is \(\frac{x^2}{4}-\frac{y^2}{5}=1\)

Question 8.
Vertices (0, ±5), foci (0, ±8)
Solution:
Vertices (0, ±5), foci (0, ±8)
Here, the vertices are on the y-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\).
Since the vertices are (0, ±5)
⇒ b = 5
Since the foci are (0, ±8)
⇒ c = 8
We know that a² + b² = c²
⇒ a² + 5² = 8²
⇒ a² = 64 – 25
⇒ a² = 39
Thus the equation of the hyperbola is \(\frac{y^2}{25}-\frac{x^2}{39}=1\)

Question 9.
Vertices (0, ±3), foci (0, ±5)
Solution:
Vertices (0, ±3), foci (0, ±5).
Here, the vertices are on the y-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\).
Since the vertices are (0, ±3)
⇒ b = 3
Since the foci are (0, ±5)
⇒ c = 5
We know that a² + b² = c²
⇒ 3² + a² = 5²
⇒ a² = 25 – 9
⇒ a² = 16
Thus, the equation of the hyperbola is \(\frac{y^2}{9}-\frac{x^2}{16}=1\).

Question 10.
Foci (±5, 0), the transverse axis is of length 8.
Solution:
Foci (±5, 0)
The transverse axis is of length 8.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
Since the foci are (±5, 0)
⇒ c = 5
Since the length of the transverse axis is 8
⇒ 2a = 8
⇒ a = 4
We know that a² + b² = c²
⇒ 4² + b² = 5²
⇒ b² = 25 – 16
⇒ b² = 9
Thus, the equation of the hyperbola is \(\frac{x^2}{16}-\frac{y^2}{9}=1\).

Question 11.
Foci (0, ±13), the conjugate axis is of length 24.
Solution:
Foci (0, ±13), the conjugate axis is of length 24.
Here, the foci are on the y-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1\)
Since the foci are (0, ±13)
⇒ c = 13
Since the length of the conjugate axis is 24,
⇒ 2b = 24
⇒ b = 12
We know that a² + b² = c²
⇒ a² + 12² = 13²
⇒ a² = 169 – 144
⇒ a² = 25
Thus, the equation of the hyperbola is \(\frac{\mathrm{x}^2}{25}-\frac{\mathrm{y}^2}{144}=-1\).

Question 12.
Foci (±3√5, 0), the latus rectum is of length 8.
Solution:
Foci (±3√5, 0), the latus rectum is of length 8.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
Since the foci are (±3√5, 0), we have c = ±3√5
Length of latus rectum = 8
⇒ \(\frac{2 \mathrm{~b}^2}{\mathrm{a}}\) = 8
⇒ b² = 4a
We know that a² + b² = c²
⇒ a² + 4a = 45
⇒ a² + 4a – 45 = 0
⇒ a² + 9a – 5a – 45 = 0
⇒ (a + 9) (a – 5) = 0
⇒ a = -9, 5
Since a is non-negative, a = 5
b² = 4a
= 4 × 5
= 20
Thus, the equation of the hyperbola is \(\frac{x^2}{25}-\frac{y^2}{20}=1\).

Question 13.
Foci (±4, 0), the latus rectum is of length 12.
Solution:
Foci (±4, 0), the latus rectum is of length 12.
Here, the foci are on the x-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
Since the foci are (±4, 0)
⇒ c = 4
Length of the latus rectum = 12
⇒ \(\frac{2 b^2}{a}\) = 12
⇒ b² = 6a
We know that a² + b² = c²
⇒ a² + 6a = 16
⇒ a² + 6a – 16 = 0
⇒ a² + 8a – 2a – 16 = 0
⇒ (a + 8) (a – 2) = 0
⇒ a = -8, 2
Since a is non-negative, a = 2,
b² = 6a
= 6 × 2
= 12
Thus, the equation of the hyperbola is \(\frac{x^2}{4}-\frac{y^2}{12}=1\).

Question 14.
Vertices (±7, 0), e = \(\frac {4}{3}\)
Solution:
Vertices (±7, 0).
Here, the vertices are on the x-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
Since the vertices are (±7, 0)
⇒ a = 7
It is given that e = \(\frac {4}{3}\)

Thus, the equation of the hyperbola is \(\frac{x^2}{49}-\frac{9 y^2}{343}=1\).

Question 15.
Foci (0, ±√10), passing through (2, 3)
Solution:
Foci (0, ±√10).
Here, the foci are on the y-axis.
Therefore, the equation of the hyperbola is of the form \(\frac{y^2}{b^2}-\frac{x^2}{a^2}=1\)
Since the foci are (0, ±√10)
⇒ c = √10
We know that a² + b² = c²
⇒ a² + b² = 10
⇒ b² = 10 – a² ……..(1)
Since the hyperbola passes through the point (2, 3),
\(\frac{9}{b^2}-\frac{4}{a^2}=1\) ……….(2)
From equations (1) and (2), we obtain
\(\frac{9}{10-a^2}-\frac{4}{a^2}=1\)
⇒ 9a² – 4(10 – a²) = a² (10 – a²)
⇒ 9a² – 40 + 4a² = 10a² – a⁴
⇒ a⁴ + 3a² – 40 = 0
⇒ a⁴ + 8a² – 5a² – 40 = 0
⇒ a² (a² + 8) – 5(a² + 8) = 0
⇒ (a² + 18) (a² – 5) = 0
⇒ a² = 5
Now b² = 10 – a²
= 10 – 5
= 5
Thus, the equation of the hyperbola is \(\frac{y^2}{5}-\frac{x^2}{5}=1\).

Question 16.
Find the equation of the hyperbola of given length of transverse axis 6, whose vertex bisects the distance between the centre and the focus.
Solution:
Suppose C is the centre, A is the vertex, and S is the focus
Given CA = AS
⇒ a = ae – a
⇒ 2a = ae
⇒ e = 2
Length of the transverse axis = 6
⇒ 2a = 6
⇒ a = 3
b² = a² (e² – 1)
= 9(4 – 1)
= 27
Equation of the hyperbola is \(\frac{x^2}{9}-\frac{y^2}{27}=1\)
⇒ 3x² – y² = 27