JEE MainMathematics/ ToughConic Sections

A hyperbola passes through the focus of the ellipse
$\frac{x^{2}}{25} + \frac{y^{2}}{4} = 1$ and its transverse and
conjugate axes coincide with the x & y-axis respectively. Also the
eccentricity of the hyperbola is the reciprocal of the eccentricity of
the ellipse. Then x -coordinate of one of the focus of the hyperbola is

Show answer and solution

Eccentricity of ellipse

= \sqrt{\frac{25 - 4}{25}} = \frac{\sqrt{21}}{5}

\therefore\ Focus\ of\ ellipse\ is(\sqrt{21},0),e_{H} = \frac{5}{\sqrt{21}}

Hence its equation is

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{a^{2}\left( \frac{25}{21} - 1 \right)} = 1

\Rightarrow \frac{x^{2}}{a^{2}} - \frac{21y^{2}}{4a^{2}} = 1

This passes through (

\sqrt{21},0) \Rightarrow a^{2} = 21

Hence equation of hyperbola is

\frac{x^{2}}{21} - \frac{y^{2}}{4} = 1\

one of the focus of hyperbola is

\left( \frac{5}{\sqrt{21}} \times \sqrt{21},0 \right) = (5,0)\

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx

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