Question

If the locus of the foot of the perpendicular drawn from centre upon any tangent to the ellipse $\frac{x^2}{40}+\frac{y^2}{10}=1$ is ${\left(x^2+y^2\right)}^2=a{x}^2+b{y}^2$ , then $\left(a-b\right)$ is equal to

• A. $10$
• B. $20$
• C. $25$
• D. $30$

Answer 1

Answer: (D)

Slope of $OP$ $=\frac{k}{h}\Rightarrow$ slope of the tangent is $\frac{-h}{k}$
Hence, the equation of tangent $PQ$ is $\frac{(y-k)}{x-h}=\frac{-h}{k}$
$\Rightarrow$ $y=-\frac{h}{k}x+\frac{h^2+k^2}{k}$
$\Rightarrow {\left(\frac{h^2+k^2}{k}\right)}^2=40{\left(\frac{-h}{k}\right)}^2+10$ condition of tangency is $c^2=a^2{m}^2+b^2$
$\Rightarrow {\left(x^2+y^2\right)}^2=40x^2+10y^2$
$\Rightarrow a-b=30$