Find the y-coordinate of all points on the curve $2x + (y + 2)^2 = 0$ where the normal line to the curve passes through the point $(-9/2, -5)$ (not on the curve).

There appears to be $3$ y-coordinates and I'm not sure how to solve this. Can anyone give me a step-by-step walkthrough on how to do this?

## Best Answer

It suffice to differentiate and we find $$2dx+2(y+2)dy=0$$ which implies $$\frac{dy}{dx}=-\frac{1}{y+2}$$ So the slope of the normal line should be inverted, and we have the equation to be $y-y_{0}=(y_{0}+2)(x-x_{0})$. If we let $y=-5$, $x=-9/2$, plug into the equation we have:

$$-5-y_{0}=(y_{0}+2)(-9/2-x_{0})$$as well as $$2x_{0}+(y_{0}+2)^{2}=0$$

The rearranged equation is a cubic since $x_{0}=\frac{1}{2}(y_{0}+2)^{2}$. I think that's where the 3 $y$ values coming from.