An ellipse $x^2+4y^2=4$ is rotated anticlockwise through a right angle in its own plane about its center An ellipse $x^2+4y^2=4$ is rotated anticlockwise through a right angle in its own plane about its center.If the locus of the point of intersection of a tangent to ellipse in its original position with the tangent at the same point of the ellipse in its new position is given by the curve
$(x^2+y^2)^2=\lambda(x^2+y^2)+\mu xy$ where $\mu$ and $\lambda$ are positive integers.Find the value of $\lambda+\mu.$

The equation of the ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$,when it is rotated anticlockwise through a right angle in its own plane about its center,then its equation becomes $\frac{x^2}{1}+\frac{y^2}{4}=1$.
Let the point of tangency be $(2\cos\theta,\sin\theta)$
So the tangent to the first ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$ is $\frac{x\cos\theta}{2}+\frac{y\sin\theta}{1}=1$
and the tangent to the second ellipse $\frac{x^2}{1}+\frac{y^2}{4}=1$ is $\frac{2x\cos\theta}{1}+\frac{y\sin\theta}{4}=1$
Eliminating $\theta $ from the two equations gives
$16x^2+4y^2=25x^2y^2$ 
But i am not correct,i dont know where i am wrong.

If we let the point of tangency be $(2\cos\theta,\sin\theta)$, then this point moves to $(-\sin\theta,2\cos\theta)$ by the rotation, at which we have to consider the tangent line of the ellipse in its new position. 
So, we have the following two lines :
$$\frac{\cos\theta}{2}x+\frac{\sin\theta}{1}y=1$$
$$\frac{-\sin\theta}{1}x+\frac{\cos\theta}{2}y=1$$
Representing $\cos\theta,\sin\theta$ by $x,y$ and using $\cos^2\theta+\sin^2\theta=1$ will give you the answer.
