$\begin{array}{1 1}(A)\;\large\frac{1}{2} \\ (B)\;\large\frac{3}{2} \\(C)\;\large\frac{1}{8} \\(D)\;\large\frac{2 }{3}\end{array} $

$y^2 = 4x$

$\implies a = 1 $

$x^2 = - 32 y$

$\implies a = -8$

tangent to the parabola $y^2=4x$ is $y = mx + \frac{a}{m} $

$\implies y = mx + \frac{1}{m} ----(1)$

tangent to the parabola $x^2 = -32 y$ is

$y = mx + am^2$

$\implies y = mx + 8m^2 ---- (2)$

Equating (1) and (2)

$mx + \frac{1}{m} = mx + 8m^2$

$\implies 8m^3 = 1$

$\therefore m = \frac{1}{2} $

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