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

$\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} $

1 Answer

$y^2 = 4x$
$\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} $
answered Apr 7, 2014 by meena.p
edited Nov 7 by priyanka.c
 

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