Question

The locus of the mid-points of the portion of the normal to the parabola $y^2=16x$ intercepted between the curve and the axis is another parabola whose latus rectum is

Answer 1

Consider the parabola $y^2=4ax$
We have to find the locus of $R(h,k)$.
Since $Q$ has ordinate $0$,
The ordinate of $P$ is $2k$.
Also, $P$ is on the curve. Then the abscissa of $P$ is $\frac{k^2}{a}$
Now, $PQ$ is normal to the curve.
Slope of tangent to the curve at any point is
$\frac{dy}{dx}=\frac{2a}{y}$
Hence, the slope of normal at point $P$ is $-\frac{k}{a}$.
Also, the slope of normal joining $P$ and $R(h,k)$ is
$\frac{2k-k}{\left(\frac{k^2}{a}\right)-h}$
Hence, comparing slopes, we get
$\frac{2k-k}{\left(\frac{k^2}{a}\right)-h}=-\frac{k}{a}$
$\Rightarrow y^2=a(x-a)$

Hence, the locus is $y^2=4(x-4)$ and the latus rectum is $4.$