# Find the equation of common tangent to the parabola $y^2=4x$ and $x^2=4y$?

$\begin{array}{1 1}(a)\;x+y=-1\\(b)\;x+2y=-2\\(c)\;y=x\\(d)\;x^2+2y=1\end{array}$

$y^2=4x$ has a tangent $y=mx+\large\frac{1}{m}$
Which is also tangent to $x^2=4y$ so this tangent passes through $x^2=4y$
$\Rightarrow x^2=4(mx+\large\frac{1}{m})$
$x^2-4mx-\large\frac{4}{m}$$=0$
Roots of this equation is same as their is one tangent hence
$B^2=4AC$
$(-4m)^2=4\times 1\times -\large\frac{4}{m}$
$16m^2=-\large\frac{-16}{m}$
$m^3=-1$
Hence $m=-1$
Equation of tangent $y=-x-1$
$x+y=-1$
Hence (a) is the correct answer.
hui like your awnser but can we do it like taking tangent of another parabola too and then equating them proportionally?? and if not then why?