Referring to the AP Inter 1st Year Maths Study Material Chapter 10 Conic Sections Exercise 10d Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Conic Sections Solutions Exercise 10d
I.
Question 1.
Find the eccentricity of hyperbola \(\frac{\mathbf{y}^2}{\mathbf{b}^2}-\frac{\mathbf{x}^2}{\mathbf{a}^2}\) = 1.
Solution:
Given hyperbola is conjugative hyperbola.
The eccentricity e is e = \(\sqrt{1+\frac{a^2}{b^2}}\)
⇒ e = \(\sqrt{\frac{b^2+a^2}{b^2}}\)
Question 2.
Find the eccentricity of rectangular hyperbola.
Solution:
Since the length of transverse axis is equal to length of the conjugative axis i.e.,
2a = 2b ⇒ a = b.
Eccentricity e = \(\sqrt{\frac{a^2+b^2}{a^2}}=\sqrt{\frac{b^2+b^2}{b^2}}=\sqrt{\frac{2 b^2}{b^2}}\) = √2
Question 3.
Find the length of conjugate axis of the hyperbola 3x2 – 4y2 = 12.
Solution:
Given hyperbola is 3x2 – 4y2 = 12.
\(\frac{3 x^2}{12}-\frac{4 y^2}{12}\) = 1
⇒ \(\frac{x^2}{4}-\frac{y^2}{3}\) = 1 which is comparing to \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
a2 = 4, b2 = 3
Since the length of the conjugative axis = 2b = 2√3
Question 4.
Find the length of transverse axis of hyperbola x2 – 4y2 = 4.
Solution:
Given hyperbola is \(\frac{x^2}{4}-\frac{y^2}{1}\) = 1
a2 = 4, b2 = 1
Since the length of the transverse axis is 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:
The given equation is \(\frac{x^2}{16}-\frac{y^2}{9}\) = 1 ⇒ \(\frac{x^2}{4^2}-\frac{y^2}{3^2}\) = 1
On comparing this equation with the standard equation of hyperbola
i.e., \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1, we have a = 4 and b = 3
We know that a2 + b2 = c2
⇒ c2 = 42 + 32
⇒ c2 = 25
∴ The coordinates of foci are (± 5, 0)
The coordinates of the vertices are (± 4, 0), e = \(\frac{c}{a}=\frac{5}{4}\)
Length of the latus rectum \(\frac{2 b^2}{a}=\frac{2 \times 9}{4}=\frac{9}{2}\)
Question 2.
\(\frac{y^2}{9}-\frac{x^2}{27}\)
Solution:
The given equation is \(\frac{y^2}{9}-\frac{x^2}{27}\) = 1
⇒ \(\frac{y^2}{3^2}-\frac{x^2}{(\sqrt{27})^2}\) = 1
On comparing with \(\frac{y^2}{a^2}-\frac{x^2}{b^2}\) = 1, we have a = 3 and b = \(\sqrt{27}\)
We know that a2 + b2 = c2
⇒ c2 = 32 + (√27)2
⇒ c2 = 9 + 27
⇒ c2 36
⇒ c = 6
The coordinates of the foci are (0, ± 6)
The coordinates of the vertices are (0, ± 3)
Eccentncity e = \(\frac{c}{a}=\frac{6}{3}\) = 2.
Length of the latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 27}{3}\) = 18.
Question 3.
9y2 – 4x2 = 36
Solution:
The given equation is 9y2 – 4x2 = 36
It can be written as \(\frac{y^2}{4}-\frac{x^2}{9}\) = 1
⇒ \(\frac{y^2}{2^2}-\frac{x^2}{3^2}\) = 1 …………(1)
On comparing equation (1) with equation of hyperbola \(\frac{y^2}{a^2}-\frac{x^2}{b^2}\) = 1
We have a = 2 and b = 3
We know that a2 + b2 = c2
⇒ c2 = 4 + 9 = 13
⇒ c = \(\sqrt{13}\)
∴ The coordinates of the foci are (0, ±\(\sqrt{13}\))
The coordinates of the vertices are (0, ± 2)
Eccentricity e = \(\frac{c}{a}=\frac{\sqrt{13}}{2}\), Length of the latus rectum
Question 4.
16x2 – 9y2 = 576
Solution:
The given equation is 16x2 – 9y2 = 576
It can be written as \(\frac{x^2}{36}-\frac{y^2}{64}\) = 1
⇒ \(\frac{x^2}{6^2}-\frac{y^2}{8^2}\) = 1
On comparing equation (1) with equation of hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1
We have a = 6 and b = 8
We know that a2 + b2 = c2
c2 = 36 + 64
⇒ c2 = 100
⇒ c = 10
∴ The coordinates of the foci are (± 10, 0)
The coordinates of the vertices are (±6, 0), Length of the 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 equation of hyperbola \(\frac{y^2}{a^2}-\frac{x^2}{b^2}\) = 1
We have a = \(\frac{6}{\sqrt{5}}\), b = 2
We know that a2 + b2 = c2
⇒ c2 = \(\frac{36}{5}\) + 4 = \(\frac{56}{5}\)
⇒ c = \(\frac{\sqrt{56}}{\sqrt{5}}\)
⇒ c = \(\frac{2 \sqrt{14}}{\sqrt{5}}\)
∴ The coordinates of the foci are (0, ± \(\frac{2 \sqrt{14}}{\sqrt{5}}\))
The coordinates of the vertices are (0, ±\(\frac{6}{\sqrt{5}}\)
Eccentricity e = \(\frac{c}{a}=\frac{\left(\frac{2 \sqrt{14}}{\sqrt{5}}\right)}{\left(\frac{6}{\sqrt{5}}\right)}=\frac{\sqrt{14}}{3}\)
Length of the latus rectum = \(\frac{2 b^2}{a}=\frac{2 \times 4}{\left(\frac{6}{\sqrt{5}}\right)}=\frac{4 \sqrt{5}}{3}\)
Question 6.
49y2 – 16x2 = 784
Solution:
The given equation is 49y2 – 16x2 = 784
It can be written as \(\frac{y^2}{16}-\frac{x^2}{49}\) = 1
⇒ \(\frac{y^2}{4^2}-\frac{x^2}{7^2}\) = 1 ………… (1)
On comparing equation (1) with the equation of hyperbola \(\frac{1}{2}\) = 1
We have a = 4 and b = 7 we known that
a2 + b2 = c2
⇒ c2 = 16 + 49
⇒ c2 = 65
⇒ c = \(\sqrt{65}\)
∴ The coordinates of the foci are (0, ±\(\frac{1}{2}\))
The coordinates of the vertices are (0, ± 4)
Eccentncity e = \(\frac{c}{a}=\frac{\sqrt{65}}{4}\), Length of the latus rectum= \(\frac{2 b^2}{a}=\frac{2 \times 49}{4}=\frac{49}{2}\).
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.
The equation of the hyperbola is of the form \(\frac{1}{2}\) = 1
Here, 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
∴ 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.
∴ The equation of the hyperbola is of the form \(\frac{y^2}{a^2}-\frac{x^2}{b^2}\) = 1.
Here, the vertices are (0, ±5), ∴ a = 5
Since, the foci are (0, ±8), ∴ C = 8
We know that c2 = a2 + b2
⇒ 82 = 52 + b2
⇒ b2 = 64 – 25
⇒ b2 = 39
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 (O, ± 3), foci (0, ± 5). Here, the vertices are on the Y-axis.
∴ The equation of the hyperbola is of the form \(\frac{y^2}{a^2}-\frac{x^2}{b^2}\) = 1
Here the vertices are (0, ±3), ∴ a = 3
Since the foci are (O, ± 5), ∴ c = 5
We know that, a2 + b2 = c2
⇒ 32 + b2 = 52
⇒ b2 = 25 – 9
⇒ b2 = 16
Equation of the hyperbola is \(\frac{y^2}{9}-\frac{x^2}{16}\) = 1
Question 10.
Foci (±5, 0), 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
∴ The equation of the hyperbola is of the form \(\frac{1}{2}\) = 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
∴ The equation of the hyperbola is \(\frac{y^2}{9}-\frac{x^2}{16}\) = 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 Y-axis.
∴ The equation of the hyperbola is of the form \(\frac{1}{2}\) = 1
Here, the foci are (0, ±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
The equation of the hyperbola is \(\frac{x^2}{25}-\frac{y^2}{20}\) = 1.
Question 12.
Foci (± 3√5, O). the lattis 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.
∴ The equation of the hyperbola is of the form \(\frac{1}{2}\) = 1
Since the foci are (±3√5, 0) ⇒ c = 3
Length of the latus recturn = 8
⇒ \(\frac{2 b^2}{a}\) = 8
⇒ b2 = 4a
We know that a2 + b2 = c2
= a2 + 4a= 45
= a2 + 4a – 45 = 0
⇒ a2 + 9a – 5a – 45 = O
⇒ (a + 9)(a – 5) = 0
⇒ a = -9, 5
Since, a is non-negative ∴ a = 5 ⇒ b2 – 4a = 4 × 5 = 20
∴ The equation of the hyperbola is \(\frac{x^2}{4}-\frac{y^2}{12}\) = 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.
∴ The equation of the hyperbola is of the form 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= 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
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:
Given that vertices (± 7, 0), e = \(\frac{4}{3}\)
Here the vertices are on the X-axis.
∴ 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 = \(\frac{4}{3} \Rightarrow \frac{c}{a}=\frac{4}{3} \Rightarrow \frac{c}{7}=\frac{4}{3}\)
c = \(\frac{28}{3}\) [∵ e = \(\frac{c}{a}\)]
We know that a2 + b2 = c2
72 + b2 = \(\left(\frac{28}{3}\right)^2\)
⇒ b2 = \(\frac{784}{9}\) – 49
⇒ b2 = \(\frac{784-441}{9}\)
⇒ b2 = \(\frac{343}{9}\)
∴ Thus the equation of the hyperbola is \(\frac{x^2}{49}-\frac{y^2}{\left(\frac{343}{9}\right)}\) = 1
⇒ \(\frac{x^2}{49}-\frac{9 y^2}{343}\) = 1
Question 15.
Foci (0, ±\(\sqrt{10}\)), passing through (2, 3).
Solution:
Foci (0, ±\(\sqrt{10}\)), passing through (2, 3). Here foci are on the Y-axis.
∴ The equation of the hyperbola is of the form – – 1
Here, the foci are (0, ± \(\sqrt{10}\)) ⇒ c = \(\sqrt{10}\)
We know that a2 + b2 = c2
⇒ a2 + b2 = 10
⇒ b2 = 10 – a2 …………….. (1)
Since the hyperbola passing through point (2, 3)
∴ \(\frac{9}{a^2}-\frac{4}{\left(10-a^2\right)}\) = 1 ………. (2)
From (1) and (2) we have
⇒ \(\frac{9}{a^2}-\frac{4}{\left(10-a^2\right)}\) = 1
9(10 – a2) – 4a2 = a2(10 – a2)
90 – 9a2 – 4a = 10a2 – a4
⇒ a4 – 23a2 + 90 = 0
⇒ a4 – 18a2 – 5a2 + 90 = 0
⇒ a2(a2 – 18) – 5(a2 – 18) = 0
⇒ (a2 -18)(a2 – 5) = 0
⇒ a2 = 18 or 5
In hyperbola c > a, i.e., c2 > a2
⇒ ∴ a2 = 5
⇒ b2 = 10 – a2
⇒ b2 = 10 – 5
⇒ b2 = 5
∴ Thus the equation of 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 bisect., the distance between the centre and the focus.
Solution:
Given 2a – 6 ⇒ a = 3 ⇒ a2 = 9.
Since the vertex bisects the distance between the
centre and focus. So c = 2a
⇒ c = 2(3) = 6
⇒ c = 6
∴ c2 = a2 + b2
⇒ b2 = c2 – a2
⇒ b2 = 62 – 32
⇒ b2 = 27
⇒ b = \(\sqrt{27}\)
⇒ b = 3√3
The equation of the hyperbola is \(\frac{x^2}{9}-\frac{y^2}{27}\) = 1.