Question

From any point R two normals which are right angled to one another are drawn to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,(a>b)$ If the feet of the normals are P and Q then the locus of the circumcentre of the triangle PQR is

• A. $\frac{x^2+y^2}{a^2-b^2}={(\frac{x^2}{a^2}+\frac{y^2}{b^2})}^2$
• B. $\frac{x^2-y^2}{a^2-b^2}={(\frac{x^2}{a^2}-\frac{y^2}{b^2})}^2$
• C. $\frac{x^2+y^2}{a^2-b^2}={(\frac{x^2}{a^2}-\frac{y^2}{b^2})}^2$
• D. $\frac{x^2+y^2}{a^2+b^2}={(\frac{x^2}{a^2}-\frac{y^2}{b^2})}^2$

Answer 1

Answer: (C)

$\frac{x^2+y^2}{a^2-b^2}={(\frac{x^2}{a^2}-\frac{y^2}{b^2})}^2$

Clearly tangent at P and Q intersect at right-angles at S (say)
$\Rightarrow$ PSQR is cyclic

$\Rightarrow$ S lies on director circle of hyperbola

$\Rightarrow S=\sqrt{a^2-b^2}\cos \theta ,\sqrt{a^2-b^2}\sin \theta$

$\Rightarrow$ Chord with middle point (h,k) i.e. circumcentre will be same as equation of chord of contact w.r.$\Rightarrow \perp$ s

$\Rightarrow \frac{xh}{a^2}-\frac{yk}{b^2}=\frac{h^2}{a^2}-\frac{k^2}{b^2}$ and $\frac{x\sqrt{a^2-b^2\cos \theta }}{a^2}-\frac{y\sqrt{a^2-b^2}\cos \theta }{b^2}=1$ are identical comparing and solving we get locus as $\frac{x^2+y^2}{a^2-b^2}={(\frac{x^2}{a^2}-\frac{y^2}{b^2})}^2$