If the eccentricity of a hyperbola (x^2)/(9)-(y^2)/(b^2)=1 , which passes through (K, 2) , is 13 3 , then the value of K^2 is

Correct answer: A. 18 Given hyperbola is (x^2)/(9)-(y^2)/(b^2)= 1 Since this passes through (K, 2) , therefore (K^2)/(9)-(4)/(b^2)= 1 Also, given e= 1+(b^2)/(a^2) = 13 3 aligned & 1+(b^2)/(9) = 13 3 9+b²=13 & b=±2 aligned Now, from eq^ n (1), we have aligned & (K^2)/(9)-(4)/(4)=1 ( b=±2) & K²=18 aligned

JEE Main 20127 MayOnline

Exam: JEE Main
Year: 2012
Date: 7 May
Mode: Online
Subject: Mathematics
Chapter: Hyperbola

JEE Main 2012 · 7 May · Online · Mathematics · Hyperbola

If the eccentricity of a hyperbola (x^2)/(9)-(y^2)/(b^2)=1 , which...

If the eccentricity of a hyperbola

\frac{x^2}{9}-\frac{y^2}{b^2}=1

, which passes through

(K, 2)

, is

\frac{\sqrt{13}}{3}

, then the value of

K^2

Options

Correct answer

A. 18

Step-by-step solution

Given hyperbola is

\frac{x^2}{9}-\frac{y^2}{b^2}=

1 Since this passes through

(K, 2)

, therefore

\frac{K^2}{9}-\frac{4}{b^2}=

1 Also, given

e=\sqrt{1+\frac{b^2}{a^2}}=\frac{\sqrt{13}}{3}

\begin{aligned}& \Rightarrow \sqrt{1+\frac{b^2}{9}}=\frac{\sqrt{13}}{3} \Rightarrow9+b²=13 \\ & \Rightarrow b=\pm2 \end{aligned}

Now,

from \mathrm{eq}^{\mathrm{n}}

(1), we have

\begin{aligned}& \frac{K^2}{9}-\frac{4}{4}=1 \quad(\because b=\pm2) \\ & \Rightarrow \quad K²=18 \end{aligned}

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