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

Question

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

Accepted Answer

Let tangent to ${y^2} = 4x$ be $$y = mx + {1 \over m}$$

Since this is also tangent to ${x^2} =  - 32y$

$$\therefore$$ $${x^2} =  - 32\left( {mx + {1 \over m}} \right)$$

$$ \Rightarrow {x^2} + 32mx + {{32} \over m} = 0$$

Now, $D=0$

$${\left( {32} \right)^2} - 4\left( {{{32} \over m}} \right) = 0$$

$$ \Rightarrow {m^3} = {4 \over {32}} \Rightarrow m = {1 \over 2}$$

Solutions · 1