Finding the Cartesian equation of an ellipse (Midpoints)

Question

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.

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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?

Answer 1

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.

Answer 2

Equation of normal at $(a\cos t,b\sin t)$ is:

$$ax\sec t-by\csc t=a^2-b^2$$

The $x,y$-intercepts are $\frac{a^2-b^2}{a} \cos t$ and $\frac{a^2-b^2}{b} \sin t$ respectively.

Hence the locus is

$$\left( \frac{a^2-b^2}{2a} \cos t,\frac{a^2-b^2}{2b} \sin t \right)$$