Gujarati
10-2. Parabola, Ellipse, Hyperbola
hard

If the straight line $x\cos \alpha + y\sin \alpha = p$ be a tangent to the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$, then

A

${a^2}{\cos ^2}\alpha + {b^2}{\sin ^2}\alpha = {p^2}$

B

${a^2}{\cos ^2}\alpha - {b^2}{\sin ^2}\alpha = {p^2}$

C

${a^2}{\sin ^2}\alpha + {b^2}{\cos ^2}\alpha = {p^2}$

D

${a^2}{\sin ^2}\alpha - {b^2}{\cos ^2}\alpha = {p^2}$

Solution

(b) $x\cos \alpha + y\sin \alpha = p$

$\Rightarrow y = – \cot \alpha .\,\,x + p{\rm{cosec }}\alpha $

It is tangent to the hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$

Therefore, ${p^2}{\rm{cose}}{{\rm{c}}^2}\alpha = {a^2}{\cot ^2}\alpha – {b^2}$

$ \Rightarrow {a^2}{\cos ^2}\alpha – {b^2}{\sin ^2}\alpha = {p^2}$.

Standard 11
Mathematics

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