<s>[INST] Finding the Cartesian equation of an ellipse (Midpoints) 

Question: The normal to the ellipse $ \frac{x^2}{25} + \frac{y^2}{9} = 1$ at a point $Q$ meets the coordinate axes at A and B respectively.
As $Q$ varies, the locus of the midpoint of $AB$ is another ellipse.
Find the Cartesian equation of this ellipse.



What I have done 
Let $x = 5 \cos(\theta)$ and $y=3\sin(\theta)$ then
$$ \frac{dx}{d\theta} = -5 \sin(\theta) , \frac{dy}{d\theta}=3cos(\theta)$$
$$ \frac{dy}{dx} = \frac{dy}{d\theta} *\frac{d\theta}{dx} $$
$$ \frac{dy}{dx} = \frac{-3\cos(\theta)}{5 \sin(\theta)}$$
$$ m_{normal} \cdot m_{tgt} = -1 $$
$$ m_{normal} = \frac{5\sin(\theta)}{3\cos(\theta)}$$
For equation of normal 
$$ y-y_1 = m(x-x_1) $$
$$ y - 3\sin(\theta)=  \frac{5\sin(\theta)}{3\cos(\theta)}(x-5\cos(\theta))$$
$$ y - 3\sin(\theta) = \frac{5\sin(\theta)}{3\cos(\theta)}x - \frac{25}{3}\sin(\theta)$$ 
$$ y= \frac{5\sin(\theta)}{3\cos(\theta)}x - \frac{16}{3} \sin(\theta) $$
At 
$$  y=0, x= \frac{16 \cos(\theta)}{5} $$
and
$$ x = 0 , y = - \frac{16}{3} \sin(\theta)$$
So we have 2 points which lie on the new ellipse
Hence the midpoint is given by $$\frac{1}{2} (X_1+X_2) $$
so 
$$ \frac{1}{2}(0 + \frac{16 \cos(\theta)}{5}) $$
Midpoint is 
$$ \frac{8\cos(\theta)}{5} $$
Now I am stuck , how should I continue?
 [/INST] HINT...You have the $x$ coordinate, now find the $y$ coordinate, and eliminate the parameter $\theta$ using $\cos^2\theta+\sin^2\theta=1$ to obtain the cartesian equation of the ellipse.
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