Sliding ellipse

Let theta be the inclination of the major axis of ellipse.Let (x1,y1)be the focus.Then the centre of ellipse will be (x1 +aecos(theta),y1+aesin(theta)) where e is eccentricity of ellipse.We know that locus of point from where perpendicular tangents are drawn to ellipse is its director circle.Hence in the given question (0,0) lies on director circle of ellipse.We also know that locus of foot of perpendicular from focus on any tangent to ellipse is its auxiliary circle.Hence in the given question points (x1,0) and (0,y1) lie on auxiliary circle of ellipse.These two informations are sufficient to eliminate theta and find the required locus.We get the locus as. $(x^2+y^2)(x^2y^2+b^4)$ = $4a^2x^2y^2$ .Hence the answer is 4a^2=100.

If the major axis of the ellipse is inclined at an angle $\theta$ to the $x$ -axis, then then coordinates of one of the two foci are $x \; = \; \sqrt{a^2\cos^2\theta + b^2\sin^2\theta} + ae\cos\theta \qquad \qquad y \; = \; \sqrt{a^2\sin^2\theta + b^2\cos^2\theta} + ae\sin\theta$ That the centre of the ellipse has coordinates $\big( \sqrt{a^2\cos^2\theta + b^2\sin^2\theta} , \sqrt{a^2\sin^2\theta + b^2\cos^2\theta} \big)$ is an elementary deduction from the fact that the origin must lie on the ellipse's director's circle. But then $\begin{array}{rcl} (x - ae\cos\theta)^2 & = & a^2\cos^2\theta + b^2\sin^2\theta \\ x^2 - 2xae\cos\theta - b^2 & = & 0 \\ \cos\theta & = & \frac{x^2 - b^2}{2aex} \end{array}$ and, similarly, $\sin\theta \; = \; \frac{y^2 - b^2}{2aey}$ so that $\begin{array}{rcl} \left(\frac{x^2 - b^2}{2aex}\right)^2 + \left(\frac{y^2 - b^2}{2aey}\right)^2 & = & 1 \\ (x^2 - b^2)^2y^2 + (y^2 - b^2)^2x^2 & = & 4a^2e^2x^2y^2 \\ x^4y^2 + x^2y^4 - 4a^2x^2y^2 + b^4x^2 + b^4y^2 & = & 0 \end{array}$ (or at least the portion of this curve for $x,y > 0$ ) is the locus of the foci of the ellipse. Both foci traverse the same locus.

For this problem, then, the answer is $4a^2 = \boxed{100}$ .