# 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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