What is the equation of common tangent to the parabola $y^2$ = 4ax and $x^2$ = 4ay ?

Solution :

The equation of tangent in slope form to $y^2$ = 4ax is

y = mx + $a\over m$

Now, if it is common to both parabola, it also lies on second parabola

then $x^2$ = 4a(mx + $a\over m$)

$mx^2 – 4am^2 – 4a^2$ = 0 has equal roots.

then its discriminant is zero. i.e. $b^2-4ac$ = 0

$16a^2m^2 + 16a^2m$ = 0

m = -1

Putting m = -1 in equation y = mx + $a\over m$ we get

y = -x – a

x + y + a = 0

Which is the required equation of common tangent to both parabola.

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