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 (-27,-50) (not on curve).
There are 3 answers.
I started by taking the derivative of the function and got: $$dy/dx=mtan=-1/(y+2)$$
So the slope of the normal is $y+2$
I then used the point in the slope=slope formula and got the following equation for the normal line $$xy+26y+2x+4$$
Which I then set equal to the original equation to find the points of intersection, and ended up with $$y^2-xy-22y$$
The general formula for the y values is x + 22, and I don't know how to get 3 values from this.
Best Answer
Hint: The line connecting $\left(-\dfrac{(y+2)^2}{2}, y\right)$ and $(-27,-50)$ has slope $y+2$: $$\frac{-(y+2)^2/2+27}{y+50} = y+2 \quad \iff \quad -(y+2)^2 + 54 = 2(y+2)(y+50).$$ This gives you two solutions. For the third one note that slope of the normal line may also be equal to $\infty$.