Question

If $C$ is the centre of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and the tangent at any point $P$ on this hyperbola meets the straight lines $bx-ay$ $=0$ and $bx+ay=0$ in the point $Q$ and $R$ respectively, then $CQ.CR$ is equal to :

• A. $a^2+b^2$
• B. $|a^2-b^2|$
• C. $\frac{b^2}{a^2}$
• D. $\frac{a^2}{b^2}$

Answer 1

Answer: (A)

$a^2+b^2$

Equation of tangent in parametric form to the given hyperbola is
$\frac{x\sec \theta }{a}+\frac{y\tan \theta }{b}=1$ (1)
It intersects $bx-ay=0$
Put $x=\frac{ay}{b}$ in equation (1) to give $Q$
$x=\frac{a(1+\sin \theta )}{\cos \theta },y=\frac{b(1+\sin \theta )}{\cos \theta }$
Similarly, for $R$
$x=\frac{a(1-\sin \theta )}{\cos \theta },y=\frac{-b(1-\sin \theta )}{\cos \theta }$
$CQ=\sqrt{{\left(\frac{b(1+\sin \theta )}{\cos \theta }\right)}^2+{\left(\frac{a(1+\sin \theta )}{\cos \theta }\right)}^2}$
$CR=\sqrt{{\left(\frac{b(1-\sin \theta )}{\cos \theta }\right)}^2+{\left(\frac{a(1-\sin \theta )}{\cos \theta }\right)}^2}$
$CQ.CR=a^2+b^2$