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KEAM 2014 Math Paper

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Question : 118 of 120

Marks: +1, -0

If the length of the latus rectum and the length of transverse axis of a hyperbola are

4\sqrt{3}

2\sqrt{3}

respectively, then the equation of the hyperbola is

$\frac{x^{2}}{3}-\frac{y^{2}}{4}=1$
$\frac{x^{2}}{3}-\frac{y^{2}}{9}=1$
$\frac{x^{2}}{6}-\frac{y^{2}}{9}=1$
$\frac{x^{2}}{6}-\frac{y^{2}}{3}=1$
$\frac{x^{2}}{3}-\frac{y^{2}}{6}=1$

Solution:

Given, length of latusrectum is

4\sqrt{3}

.

\therefore \;\; \;\frac{2b^{2}}{a}=4\sqrt{3}

....(i)

and length of transverse axis is

2\sqrt{3}

.

\therefore \;\; 2a=2\sqrt{3} \Rightarrow a=\sqrt{3}

On putting

a=\sqrt{3}

in Eq. (i), we get

\;\frac{2b^{2}}{\sqrt{3}}=4\sqrt{3} \Rightarrow 2b^{2}=12 \Rightarrow b^{2}=6

\therefore

Equation of hyperbola is

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

\therefore \;\; \;\frac{x^{2}}{3}-\;\frac{y^{2}}{6}=1

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