If the chord of the locus of the perpendicular drawn from centre upon any tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $(x^2 + y^2)^2 = a^2x^2 + b^2y^2$, then $(a - b)$ is equal to
Step-by-Step Solution
Key Concept: The tangent to an ellipse at a point has slope equal to the slope of the radius at that point, and tangency conditions constrain the ellipse parameters.
The slope of $OP$ equals the slope of the tangent at point $P$, which is $\frac{k}{h}$. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the tangent at $(h,k)$ is $\frac{hx}{a^2} + \frac{ky}{b^2} = 1$. The tangent $PQ$ has equation $y = -\frac{h}{k}x + \frac{a^2+b^2}{k}$. Using the condition of tangency $c^2 = a^2m^2 + b^2$ where $m = -\frac{h}{k}$, we get $(x^2+y^2)^2 = 40z^2 + 10y^2$, which simplifies to $a^2 - b^2 = 30$.
Correct Answer: 30