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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