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 $\left(x^{2} + y^{2}\right)^{2}=ax^{2}+by^{2}$ , 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 $\frac{\left(\right. y - k \left.\right)}{x - h}=\frac{- h}{k}$
$\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} $
$\Rightarrow \left(x^{2} + y^{2}\right)^{2}=40x^{2}+10y^{2}$
$\Rightarrow a-b=30$