Question

Two equal parabolas with axes in opposite directions touch at a point O. From a point P on one of them are drawn tangents PQ and PQ' to the other. Prove that QQ' will touch the first parabola in P' where PP' is parallel to the common tangent at O.

Answer 1

Let the two parabolas be

$y^2=4ax.......\left(i\right)y^2=-4ax......\left(ii\right)$

Let tangents are drawn from a point $P(a{t}^2,2at)$ on $\left(i\right)$ to $\left(ii\right)$

Then $Q{Q}^{\prime }$ is a chord of contact to $\left(ii\right)$ w.r.t. to $P$

Equation of chord of contact is $T=0$

$y{y}_1=-2a(x+x_1)2aty=-2a(x+a{t}^2)ty=-x-a{t}^2......\left(iii\right)$

substituting $x$ from $\left(iii\right)$ in $\left(i\right)$

$y^2=4a(-a{t}^2-ty)y^2+4aty+4a^2{t}^2=0{(y+2at)}^2=0\Rightarrow y=-2at$

We are getting equal roots from the equation. So, $Q{Q}^{\prime }$ touch the parabola

Substituting $y=-2at$ in $\left(i\right)$

${(-2at)}^2=4ax\Rightarrow x=a{t}^2$

So the point of contact to $\left(i\right)$ is $P^{\prime }(a{t}^2,-2at)$

Equation of common tangent at $O$ is

$x=0$

Slope of $P{P}^{\prime }$$=\frac{-2at-2at}{a{t}^2-a{t}^2}=\infty$

So, $P{P}^{\prime }$ is perpendicular to $x$ axis

Hence it is parallel to the $y$ axis or $x=0$

Hence proved.