The eccentricity of the hyperbola whose transverse axis is $2 a$ having coordinate axes as its axes and passing through the point $(h, k)$ is

\sqrt{\frac{h^{2} + k^{2} - a^{2}}{k^{2} + a^{2}}}

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\sqrt{\frac{h^{2} + k^{2} - a^{2}}{h^{2} - a^{2}}}

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\sqrt{\frac{h^{2} - k^{2} + a^{2}}{k^{2} - a^{2}}}

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\sqrt{\frac{h^{2} - k^{2} - a^{2}}{h^{2} - a^{2}}}

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Solution

The correct option is C $\sqrt{\frac{h^{2} + k^{2} - a^{2}}{h^{2} - a^{2}}}$
Let eccentricity be $e$
Foci are $(a e, 0)$ and $(- a e, 0)$
$\sqrt{(h - a e)^{2} + k^{2}} - \sqrt{(h + a e)^{2} + k^{2}} = 2 a$
$2 (h^{2} + a^{2} e^{2} + k^{2}) + 2 \sqrt{((h + a e)^{2} + k^{2}) ((h - a e^{2}) + k^{2})} = 4 a^{2}$
$\sqrt{((h + a e)^{2} + k^{2}) ((h - a e)^{2} + k^{2})} = 2 a^{2} - h^{2} - a^{2} e^{2} - k^{2}$
Squaring on both sides, we get
$\Rightarrow (h^{2} - a^{2}) e^{2} = h^{2} + k^{2} - a^{2}$
$e = \sqrt{\frac{h^{2} + k^{2} - a^{2}}{h^{2} - a^{2}}}$

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Similar questions

T a n g e n t s a r e f r o m t h e p o i n t (h, k) t o t h e c i r c l e x^{2} + y^{2} = a^{2}; t h e n t h e t h e a r e a o f t h e t r i a n g l e f o r m e d b y t h e s t r a i g h t l i n e j o i n i n g t h e i r p o i n t s o f c o n t a c t i s \frac{a {(h^{2} + k^{2} - a^{2})}^{\frac{3}{2}}}{h^{2} + k^{2}} .

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

h

k

\frac{a^{2}}{h^{2}} + \frac{b^{2}}{k^{2}} =

h^{2} + k^{2} > a^{2}

h x + k y = a^{2}

x^{2} + y^{2} = a^{2}

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

h

k

\frac{a^{2}}{h^{2}} + \frac{b^{2}}{k^{2}} =

x^{2} + y^{2} = a^{2}

h^{2} + k^{2} - a^{2} < 0

h x + k y = a^{2}

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