Question

The polar of $c^2{x}^2+d^2{y}^2=1$ is tangent to$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ then locus of pole of $c^2{x}^2+d^2{y}^2=1$ is :

• A. $a^2{c}^4{x}^2=b^2{d}^4{y}^2$
• B. $a^2{c}^4{x}^2+b^2{d}^4{y}^2=1$
• C. $a{c}^2{x}^2+b{d}^2{y}^2=1$
• D. $ax+by=\frac{a^2+b^2}{c^2+d^2}$

Answer 1

The correct option is B $a^2{c}^4{x}^2+b^2{d}^4{y}^2=1$

Let the pole be $(h,k)$
Then equation of polar is $c^{2}hx+d^{2}ky=1$
Since, it is tangent to $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Therefore, $(intercept{)}^2=a^2{m}^2+b^2$
$\Rightarrow \frac{1}{d^4{k}^2}=\frac{a^2{h}^2{c}^4}{d^4{k}^2}+b^2$
$\Rightarrow a^2{c}^4{h}^2+b^2{d}^4{k}^2=1$