Triangles are formed by pair of tangents drawn from any point on ellipse. Prove that the orthocentre of each triangle lies on the ellipse 
A triangle is formed by pair of tangents drawn from a point on the ellipse $$a^2x^2+b^2y^2=(a^2+b^2)^2$$ 
  to the ellipse $$b^2x^2+a^2y^2=a^2b^2$$ and their chord of contact. 
Find the locus of orthocentre of triangle 

 A: HINT.-MvG's answer tells you that the calculations can be long and apparently complicated. I give you here a mode that may seem shorter and easier for you.
Let $E_1$ and $E_2$ the two ellipses so$$E_1:\frac{x^2}{A^2}+\frac{y^2}{B^2}=1;\space A=\frac{a^2+b^2}{a},\space B=\frac{a^2+b^2}{b}\\E_2: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ It is convenient to start from two points $P(x_1,y_1)$ and $Q(x_2,y_2)$  of $E_2$. The corresponding tangents are
$$T_1:\frac{x_1X}{a^2}+\frac{y_1Y}{b^2}=1\text{ with pente }-\frac{x_1b^2}{y_1a^2}\\T_2:\frac{x_2X}{a^2}+\frac{y_2Y}{b^2}=1\text{ with pente }-\frac{x_2b^2}{y_2a^2}$$ $T_1$ and $T_2$ go through the point $T(x_0,y_0)\in E_1$ so we have
$$\frac{x_1x_0}{a^2}+\frac{y_1y_0}{b^2}=\frac{x_2x_0}{a^2}+\frac{y_2y_0}{b^2}\iff x_0\frac{x_1-x_2}{a^2}=y_0\frac{y_2-y_1}{b^2}\space (*)$$
The perpendicular from $P(x_1,y_1)$ to the line $QT$ has equation
$$(Y-y_1)=(X-x_1)\frac{y_2a^2}{x_2b^2}$$ similarly we have
$$(Y-y_2)=(X-x_2)\frac{y_1a^2}{x_1b^2}$$ These two perpendiculars intersect at the orthocentre $H$ of the post and this common point should belong to the elipse $E_2$ which can be verified using the relation $(*)$ above and the coordinates of the found orthocentre (keep in mind that $\frac{x_0^2}{A^2}+\frac{y_0^2}{B^2}=1$).
I leave to you the (non hard) final calculation.
