Example 4: Find the curves that are perpendicular to the parabolas in Example 3.

It is said:

$\displaystyle \frac{dy}{dx}=\frac{-1}{slopeofparabola}=\frac{-x}{2y}$

My question is about $\displaystyle \frac{-x}{2y}$ and why it is so.

The equations for the parabolas in Example 3 are:

$\displaystyle x^2$

$\displaystyle 2x^2$

$\displaystyle -x^2$

$\displaystyle 0x^2$

$\displaystyle y=-2y^2$

$\displaystyle 100x^2$

I read through Example 3 and I understand to the extent that I follow the arithmetic and I see how the equations for the parabolas were derived.

The $\displaystyle \frac{-x}{2y}$ in $\displaystyle \frac{dy}{dx}=\frac{-1}{slopeofparabola}=\frac{-x}{2y}$ I am having some trouble understanding why that is or what is really being said. I don't think I've really been through this topic of a curve being perpendicular to another curve before.

Thank You...