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Let \mathrm{\lambda x-2y= \mu } be a tangent to the hyperbola \mathrm{a^{2}x^{2}-y^{2}= b^{2}}. Then \mathrm{\left ( \frac{\lambda }{a} \right )^{2}-\left ( \frac{\mu}{b} \right )^{2}} is equal to:

Option: 1

-2


Option: 2

-4


Option: 3

2


Option: 4

4


Answers (1)

best_answer

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

\mathrm{Tangent: \quad y= \frac{\lambda\, }{2}x-\frac{\mu}{2}}

Using condition for tangency for hyperbola
\mathrm{\frac{\mu^{2}}{4}= \frac{b^{2}}{a^{2}}\times\frac{\lambda^{2}}{4}-b^{2}}
\mathrm{\frac{\mu^{2}}{4b^{2}}= \frac{1}{4}\left ( \frac{\lambda^{2}}{a^{2}}-4 \right )}
\mathrm{\frac{\lambda^{2}}{a^{2}}- \frac{\mu^{2}}{b^{2}}= 4}

The correct answer is option (D)

Posted by

Devendra Khairwa

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