# Calculus: Normal line to curve through point not on curve

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).

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.

• Call the point of "normalcy" $(a,b)$, not $(x,y)$. – André Nicolas Oct 26 '13 at 23:11

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 $\infty$.

You have the correct expression for the slope of the normal line. You then need to state that at the point $(x_1, y_1)$ on the original curve, the slope of the normal line will be given by:$$m=y_1+2\tag{1}$$and the point $(x_1,y_1)$ will satisfy the original equation, i.e.:$$2x_1+(y_1+2)^2=0$$This leads to:$$x_1=-\frac{(y_1+2)^2}{2}\tag{2}$$So the equation of the normal line will be given by:$$y-y_1=m(x-x_1)$$Now use equations (1) and (2) to substitute $m$ and $x_1$, then use the fact that the line passes through $(-27,-50)$ to solve for $y_1$. That should give you your 3 values.

There is a similar problem here, so I have submitted a solution for this one at that location (the three points aren't as awful to solve for as the situation appears)...