The slope of the line touching both the parab | vedclass

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(C) The equation of the first parabola is $y^2 = 4x$,which is of the form $y^2 = 4ax$ with $a = 1$.The equation of the tangent to $y^2 = 4x$ with slop

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$\frac{1}{2}$

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(C) The equation of the first parabola is $y^2 = 4x$,which is of the form $y^2 = 4ax$ with $a = 1$.The equation of the tangent to $y^2 = 4x$ with slope $m$ is $y = mx + \frac{a}{m} = mx + \frac{1}{m} \quad (1)$.The equation of the second parabola is $x^2 = -32y$,which is of the form $x^2 = 4Ay$ with $4A = -32$,so $A = -8$.The equation of the tangent to $x^2 = 4Ay$ with slope $m$ is $y = mx - Am^2$. Substituting $A = -8$,we get $y = mx - (-8)m^2 = mx + 8m^2 \quad (2)$.Since the line is common to both parabolas,we equate the intercepts from $(1)$ and $(2)$:$\frac{1}{m} = 8m^2$.This simplifies to $8m^3 = 1$,which gives $m^3 = \frac{1}{8}$.Taking the cube root,we get $m = \frac{1}{2}$.

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

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