Question

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

Answer 1

Compute the slope of the line touches parabola:

${\mathrm{x}}^2=-32\mathrm{y},{\mathrm{y}}^2=4\mathrm{x}$
$\mathrm{m}$ be the slope of the common tangent
Equation of tangent of parabola $y^2=4ax$

$\mathbf{y}\mathbf{=}\mathbf{mx}\mathbf{+}\frac{\mathbf{a}}{\mathbf{m}}$

Here, $a=1$

$y=mx+\frac{1}{m}...\left(i\right)$
Equation of tangent of parabola $x^2=-4ay$
$\mathbf{y}\mathbf{=}\mathbf{mx}\mathbf{+}{\mathbf{am}}^{\mathbf{2}}\mathbf{}$

$\mathrm{y}=\mathrm{mx}+8{\mathrm{m}}^2...\left(ii\right)$
$\left(i\right)and\left(ii\right)$ are identical

$\begin{array}{c}\frac{1}{m}=8m^2\\ \Rightarrow m^3=\frac{1}{8}\\ m=\frac{1}{2}\end{array}$

Alternate method for computing the slope of the line

Let tangent to$y^2=4x$ be $y=mx+\frac{1}{m}$
as this is also tangent to $x^2=-32y$
Solving $x^2+32mx+\frac{32}{m}=0$
Since roots are equal

$\begin{array}{rl}& \therefore D=0\\ & \Rightarrow (32{)}^2-4\times \frac{32}{m}=0\\ & \Rightarrow m^3=\frac{4}{32}\\ & \Rightarrow m=\frac{1}{2}\end{array}$

The slope of the line touching both the parabolas is $\mathit{m}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$

Hence, the slope of line touching both the parabolas is $\mathit{m}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}$