Animated Solution for Mathematics - Conic Sections: The slope of the line touching both the parabolas $y^2=4x$ and $x^2=-32y$ is

Imagine you are standing on a coordinate plane. To your right, the parabola $y^2=4x$ opens its arms wide, embracing the positive $x$-axis. Below you, the parabola $x^2=-32y$ stretches downwards, a mirror image of growth in the negative direction.

Our mission is to find a single, straight line that acts as a bridge, perfectly grazing both of these curves. This is the common tangent.

When we face a problem like this, we utilize the slope form of a tangent. For any parabola of the form $y^2=4ax$, the equation of a tangent with a slope $m$ is given by the elegant formula:

By comparing our parabola $y^2=4x$ with the standard form $y^2=4ax$, we identify that $4a=4$, which means $a=1$. Thus, our tangent line is simply:

Substitution and the Discriminant

For this line to be a common tangent, it must also touch the second parabola, $x^2=-32y$. We enforce this by substituting our expression for $y$ into the equation of the second parabola:

This simplifies to $x^2=-32mx-\frac{32}{m}$. Rearranging this into a standard quadratic form, we obtain:

If this line is a tangent to the second parabola, it must touch the curve at exactly one point. In the language of algebra, this means our quadratic equation must have equal roots, implying the discriminant $D=b^2-4ac$ must be exactly zero.

In our equation, $a=1$, $b=32m$, and $c=\frac{32}{m}$. Setting the discriminant to zero:

Multiplying by $m$ (given $m eq 0$), we arrive at $1024m^3-128=0$. This simplifies to:

Taking the cube root, we find the slope of the common tangent is $m=\frac{1}{2}$.