Question

The perpendicular from the centre upon the normal on any point of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ meets at $R$. Find the locus of $R$.

• A. $(a^2{y}^2-b^2{x}^2)(x^2+y^2{)}^2=(a^2+b^2{)}^2{x}^2{y}^2$
• B. $(a^2{y}^2-b^2{x}^2)(x^2+y^2)=(a^4+b^4)x^2{y}^2$
• C. $(a^2{y}^2-b^2{x}^2)(2x^2+y^2)=(a^2+2b^2)x^2{y}^2$
• D. None of these

Answer 1

Answer: (A)

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

Equation of normal to the given hyperbola is given by,
$\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }=a^2+b^2..\left(1\right)$
Slope $m=-\frac{a\tan \theta }{b\sec \theta }=-\frac{a\sin \theta }{b}$
Thus equation of line through centre and perpendicular to the (1) is, $y=-\frac{1}{m}x$
$\Rightarrow y=\frac{b}{a\sin \theta }x..\left(2\right)$
Thus by eliminating $\theta$ from (1) and (2) we will get required locus,
$\Rightarrow \cos \theta (ax+\frac{by}{bx/ay})=a^2+b^2$
$\Rightarrow a\cos \theta (x^2+y^2)=x(a^2+b^2)$
Squaring we get,
$\Rightarrow a^2{\cos }^2\theta {(x^2+y^2)}^2=x^2{(a^2+b^2)}^2$
$\Rightarrow a^2(1-\frac{b^2{x}^2}{a^2{y}^2}){(x^2+y^2)}^2=x^2{(a^2+b^2)}^2$
$\Rightarrow (x^2+y^2{)}^2(a^2{y}^2-b^2{x}^2)=x^2{y}^2(a^2+b^2{)}^2$