Question

Two parabolas have a common axis and concavities in opposite directions ; if any line parallel to the common axis meet the parabolas in $P$ and $P^{\prime }$, prove that the locus of the middle point of $P{P}^{\prime }$ is another parabola, provided that the latera recta of the given parabolas are unequal.

Answer 1

Let the two unequal parabolas be $y^2=4ax.....\left(i\right)$ and $y^2=-4bx......\left(ii\right)$

Let the mid point of $P{P}^{\prime }$ be $(h,k)$

Then the line parallel to axis is $y=k$

substitute $y=k$ in $\left(i\right)$

$k^2=4ax\Rightarrow x=\frac{k^2}{4a}$

So the point $P$ is $(\frac{k^2}{4a},k)$

substituite $y=k$ in $\left(ii\right)$

$k^2=-4bx\Rightarrow x=\frac{k^2}{-4b}$

So the point $P^{\prime }$ is $(-\frac{k^2}{4b},k)$

Mid point of $P{P}^{\prime }$ is

$(\frac{\frac{k^2}{4a}-\frac{k^2}{4b}}{2},\frac{k+k}{2})(\frac{k^2}{8a}-\frac{k^2}{8b},k)$

But we considerd mid point as $(h,k)$

$\Rightarrow h=\frac{k^2}{8a}-\frac{k^2}{8b}h=\frac{k^2}{8}\left(\frac{b-a}{ab}\right)k^2=\frac{8ab}{b-a}h$

Replacing $h$ by $x$ and $k$ by $y$

$y^2=\frac{8ab}{b-a}x$

clearly this represents a parabola

Hence proved.