Find the locus of intersection of tangents to an ellipse if the lines joining the points of contact to the centre be perpendicular. Find the locus of intersection of tangents to an ellipse if the lines joining the points of contact to the centre be perpendicular.
Let the equation to the tangent be $$y=mx+\sqrt{a^2m^2+b^2} $$ This has to roots for $m$ that i.e $m_1$ and $m_2$.
perpendicular to this line passing through $(0,0)$ is $$my+x=0 $$ slope is $\frac{-1}{m}$ 
so for perpendiculars $$\frac{(-1)(-1)}{m_1m_2} =-1 $$ so $m_1m_2=-1 $ 
from equation of tangent $y=mx+\sqrt{a^2m^2+b^2} $
$$m^2(x^2-a^2)-2mxy+y^2-b^2=0 $$and hence locus is $$\frac{y^2-b^2}{x^2-a^2}=-1 $$ But this is not the correct answer. The answer is $b^4x^2+a^4y^2=a^2b^2(a^2+b^2) $. What's the error ?

Let $(x_1,y_1)$ be the generic point of the locus.
It is well known that $$\frac {xx_1}{a^2}+\frac {yy_1}{b^2}=1$$ represents the line passing through the points of contact of the tangents from $(x_1,y_1)$.
Now consider the equation $$\frac {x^2}{a^2}+\frac {y^2}{b^2}-\left(\frac {xx_1}{a^2}+\frac {yy_1}{b^2}\right)^2=0$$ It is satisfied by the coordinates of the center and the points of contact.
Since it can be written $$x^2 \left(\frac {x_1^2}{a^4}-\frac 1{a^2}\right) + y^2 \left(\frac {y_1^2}{b^4}-\frac 1{b^2}\right) + \frac {2\,x\,y\,x_1y_1}{a^2b^2}=0$$
it is quadratic homogeneous so represents a pair of lines (degenerate conic)
, clearly the lines joining the points of contact to the centre.
It is not difficult to prove that the lines are mutually perpendicular iff the sum of the coefficients of $x^2$ and $y^2$ is zero, that is $$\frac {x_1^2}{a^4}+\frac {y_1^2}{b^4}=\frac 1{a^2}+\frac 1{b^2}$$
