PARABOLA : Problem Find the equation of line touching both the parabolas 
$$ x^2=-32y.......(1)$$ $$ y^2=4x.........(2) $$
i have equated slopes of both the parabolas and applied the condition that all the points on the line lie outside the parabolas.

 A: HINT:
Parametric equation of $x^2=-32y: x=8t,y=-2t^2$
The equation of tangent at $(8t,-2t^2)$  will be  $$x(8t)=-16(y-2t^2)\iff xt+2y-4t^2=0$$
Similarly find the equation of tangent of $y^2=4x$ at $(v^2,2v)$
These two equation must be same$\implies$
the ratio of the coefficients of $x$
$=$ the ratio of the coefficients of $y$
$=$  the ratio of the constants
A: General equation for a tangent of the parabola $y^2=4ax$ is $y=mx+\frac{a}{m}$.And similarly for the parabola $x^2=4by$ the general equation of tangent is $x=ny+\frac{b}{n}$.According to the question $a=1$ and $b=-8$.And by comparing the equations $y=mx+\frac{1}{m}$ and $x=ny-\frac{8}{n}$(which have to be the same) we can conclude that $m=0.5$ and $n=2$.Therefore the equation of the tangent is :- $$x=2y-4$$I hope this was helpful.
A: Let the common tangent line be $y=mx+q$. Then we know that the resolvent equations of the systems
$$
\begin{cases}
y=mx+q\\
x^2=-32y
\end{cases}
\qquad
\begin{cases}
y=mx+q\\
y^2=4x
\end{cases}
$$
have zero discriminant. The resolvent equations are
$$
x^2+32mx+32q=0
\qquad
m^2x^2+(2mq-4)x+q^2=0
$$
so we get
$$
\begin{cases}
32^2m^2-4\cdot32q=0\\[4px]
(2mq-4)^2-4m^2q^2=0
\end{cases}
$$
that simplifies to
$$
\begin{cases}
q=8m^2 \\[4px]
mq-1=0
\end{cases}
$$
Can you finish?
