Finding the locus of midpoint of $AB$ The normal to the ellipse $b^2x^2+a^2y^2=a^2b^2$  is passing through the x-axis in point $A$  and through the Y-axis in point $B$.
Point $P$ is the midpoint of $AB$.
Need to find the locus of $P$.
My attempt :I tried to use the fact of the tangent to the ellipse $n^2=a^2m^2+b^2$ and to use slopes but I didn't succeed  getting the locus. 
Thanks. 
 A: Hint.
$x(t)= a \cos t$, $y(t)=b \sin t$ is a parameterization of your ellipse.
Then for $t \in [0, 2 \pi]$, find the normal to the ellipse at point $(x(t),y(t))$ and the intersection points with the coordinates axis. Then you'll get the midpoint as a function of $t$.
A: i will parametrize the ellipse by $$x = a \cos t, y = b \sin t $$ the slope of the tangent is $$\frac{dy}{dx} = -\frac{b\cos t}{a \sin t} $$ therefore the slope of the normal is $$\frac{a\sin t}{b \cos t}.$$  the equation of the normal line is $$(y-b\sin t)b \cos t = a \sin t(x  -a \cos t) $$ the $x$-intercept is given by $$a \sin t(x - a \cos t) = -b^2\sin t \cos t \to x = \frac 1a(a^2-b^2)\cos t.$$ in the same way the $y$-intercept is $$y = \frac1b(b^2-a^2)\sin t.$$ therefore the midpoint is $$x = \frac1{2a}(a^2 - b^2)\cos t,\\y = \frac1{2b}(b^2 - a^2)\sin t $$ you can eliminate $t,$  and find that locus of the center of the two intercepts is the ellipse $$4a^2x^2 + 4b^2y^2 = (a^2-b^2)^2.$$ 
A: Hint #2.
If you don't know the parametric approach, you can use the equation $b^2x x_0+ a^2 yy_0 =a^2 b^2$ of the tangent to the ellipse at a point $(x_0,y_0)$ of the ellipse to find then the equation of the normal at the same point.
