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JEE Advanced 2006

Exam: JEE Advanced
Year: 2006
Subject: Mathematics
Chapter: Hyperbola

JEE Advanced 2006 · Mathematics · Hyperbola

If e_1 is the eccentricity of the ellipse (x^2)/(16)+(y^2)/(25)=1...

If

e_1

is the eccentricity of the ellipse

\frac{x^2}{16}+\frac{y^2}{25}=1

and

e_2

is the eccentricity of the hyperbola passing through the focii of the ellipse and

e_1 e_2=1

, then equation of the hyperbola is

Options

A $\frac{x^2}{9}-\frac{y^2}{16}=1$
B $\frac{x^2}{16}-\frac{y^2}{9}=-1$
C $\frac{x^2}{9}-\frac{y^2}{25}=1$
DNone of these

Correct answer

B. $\frac{x^2}{16}-\frac{y^2}{9}=-1$

Step-by-step solution

The eccentricity of

\frac{x^2}{16}+\frac{y^2}{25}=1

is

e_1=\sqrt{1-\frac{16}{25}}=\frac{3}{5}

\therefore \quad e_2=\frac{5}{3}

\Rightarrow

Foci of ellipse

(0, \pm 3)

\therefore

Equation of hyperbola is

\frac{x^2}{16}-\frac{y^2}{9}=-1

Hence (b) is the correct answer.

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