Question

The locus of the foot of perpendicular from the centre upon any normal to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

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

Answer 1

The correct option is B $(x^2+y^2{)}^2(a^2{y}^2-b^2{x}^2)=(a^2+b^2{)}^2{x}^2{y}^2$

The equation of any normal at $P\left(\phi \right)$ to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is
$\left(a\cos \phi \right)x+\left(b\cot \phi \right)y=a^2+b^2$

Let locus of the foot of perpendicular from origin be $(h,k)$
$\Rightarrow \frac{h-0}{a\cos \phi }=\frac{k-0}{b\cot \phi }=-\left(\frac{a^2+b^2}{a^2{\cos }^2\phi +b^2{\cot }^2\phi }\right)$
Let $\frac{h}{a\cos \phi }=\frac{k}{b\cot \phi }=-\left(\frac{a^2+b^2}{a^2{\cos }^2\phi +b^2{\cot }^2\phi }\right)=t$
$\Rightarrow h=\left(a\cos \phi \right)t$
$\Rightarrow a\cos \phi =\frac{h}{t}\cdots \left(1\right)$
$b\cot \phi =\frac{k}{t}\cdots \left(2\right)$
Now, $-\left(\frac{a^2+b^2}{a^2{\cos }^2\phi +b^2{\cot }^2\phi }\right)=t$

$\Rightarrow -\left(\frac{a^2+b^2}{\frac{h^2}{t^2}+\frac{k^2}{t^2}}\right)=t$
$\Rightarrow t=-\left(\frac{h^2+k^2}{a^2+b^2}\right)$
Eliminating $\phi$ from $\left(1\right)$ and $\left(2\right)$, we get
${\sec }^2\phi -{\tan }^2\phi =1\Rightarrow {\left(\frac{at}{h}\right)}^2-{\left(\frac{bt}{k}\right)}^2=1$
$\Rightarrow (\frac{a^2}{h^2}-\frac{b^2}{k^2})\cdot {\left(\frac{h^2+k^2}{a^2+b^2}\right)}^2=1$

$\therefore \text{Required locus is}(x^2+y^2{)}^2(a^2{y}^2-b^2{x}^2)=(a^2+b^2{)}^2{x}^2{y}^2$