If any point on a hyperbola is (3 θ, 2 θ) , then what is the eccentricity of the hyperbola?

Correct answer: D. 13 2 Let equation of hyperbola is (x^2)/(a^2)-(y^2)/(b^2)=1 then eccentricity e= 1+(a^2)/(b^2) Now, as point (3 θ, 2 θ) lies on hyperbola then, (3^2 ^2 θ)/(a^2)-(2^2 ^2 θ)/(b^2)=1(9 ^2 θ)/(a^2)-(4 ^2 θ)/(b^2)=1 This is true for a^2=9 and b^2=4 eccentricity (e)= 1+(9)/(4) = 13 2

NDA 2021

Exam: NDA
Year: 2021
Subject: Mathematics
Chapter: Hyperbola

NDA 2021 · Mathematics · Hyperbola

If any point on a hyperbola is (3 θ, 2 θ) , then what is the...

If any point on a hyperbola is

(3 \tan \theta, 2 \sec \theta)

, then what is the eccentricity of the hyperbola?

Options

A $\frac{3}{2}$
B $\frac{5}{2}$
C $\frac{\sqrt{11}}{2}$
D $\frac{\sqrt{13}}{2}$

Correct answer

D. $\frac{\sqrt{13}}{2}$

Step-by-step solution

Let equation of hyperbola is

\frac{x^2}{a^2}-\frac{y^2}{b^2}=1

then eccentricity

e=\sqrt{1+\frac{a^2}{b^2}}

Now, as point

(3 \tan \theta, 2 \sec \theta)

lies on hyperbola then,

\frac{3^2 \tan ^2 \theta}{a^2}-\frac{2^2 \sec ^2 \theta}{b^2}=1\frac{9 \tan ^2 \theta}{a^2}-\frac{4 \sec ^2 \theta}{b^2}=1

This is true for

a^2=9

and

b^2=4\therefore \text{ eccentricity }(e)=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}

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