# The slope of the line touching both the parabolas $$y^2$$ = 4x and $$x^2$$ = -32 is

## Solution :

for parabola, $$y^2$$ = 4x

Let y = mx + $$1\over m$$ is tangent line and it touches the parabola $$x^2$$ = -32.

$$\therefore$$ $$x^2$$ = -32(mx + $$1\over m$$)

$$\implies$$ $$x^2 + 32mx + {32\over m}$$ = 0

Now, D = 0 because it touches the curve.

$$\therefore$$ $$(32m)^2 – 4.{32\over m}$$ = 0

$$\implies$$ $$m^3$$ = $$1\over 8$$

$$\implies$$  m = $$1\over 2$$

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