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?
On
I'm picking up this old problem (and its "partner"), in part to correct some typos and mis-statements, and in part because it turns out to be less intimidating to solve exactly than it appears.
I'm going to call the external point $ \ (X, Y) \ , \ $ and $ \ (x_0, y_0) \ $ the point(s) we are interested in locating on the "horizontal parabola" $ \ -2x \ = \ (y+2)^2 \ $ [which has its vertex at $ \ ( 0, -2 ) \ $ and "opens to the left"] . We'll use the already established fact that the slope of the normal line at $ \ (x_0, y_0) \ $ is $ \ m \ = \ y_0 + 2 \ $ to write the equation of any such normal line through the external point as
$$ \ y_0 - Y \ = \ (y_0 + 2) \ (x_0 - X) \ \ \Rightarrow \ \ y_0 - Y \ = \ (y_0 + 2) \ (\ [-\frac{1}{2}(y_0+2)^2] - X) \ \ , $$
correcting a "typo" in user32240's answer. It is noted correctly that this leads to a cubic polynomial in $ \ y_0 \ $ of the type for which many of us have learned to salivate "Cardano!" on sight.
I will confess that I graphed the function first to see whether this looked like anything could be done without bringing in "heavier computational machinery". As it happens, because of the choices of values in the two related problems, we can do something simpler. If we re-write the left-hand side of the equation and rearrange terms, we obtain
$$ ( y_0 + 2 ) - ( Y + 2 ) \ = \ -\frac{1}{2}(y_0+2)^3 \ - \ X \ (y_0 + 2) $$
$$ \Rightarrow \ \ -\frac{1}{2}(y_0+2)^3 \ - \ (X + 1) \ (y_0 + 2) \ + \ (Y + 2) \ = \ 0 \ \ . $$
We'll multiply through by $ \ -2 \ $ to make this a monic polynomial and make the substitution $ \ u \ = \ y_0 \ + 2 \ $ to produce $ \ u^3 \ + \ 2 \ (X + 1) \ u \ - \ 2 \ (Y + 2) \ = \ 0 \ \ . $
Now we'll set to work solving the problems for the given "external" points (I use quotes for a reason we'll see shortly). For the other problem (#540862), we have $ \ (X, Y) \ = \ ( -27, -50 ) $ , for which the polynomial is $ \ u^3 \ - \ 52 u \ + \ 96 \ = \ 0 \ . $ A little work with the Rational Zeroes Theorem and synthetic division leads us to find one root is $ \ u = 2 \ $ , and the polynomial factors nicely as $ \ (u + 8) \ (u - 2) \ (u - 6) \ = \ 0 \ . $ The three points we seek are thus $ \ (-32 , -10 ) \ , \ (-2 , 0 ) \ , \ $ and $ \ (-18 , 4 ) \ . $ [This is also in response to a remark in njguliyev's answer for that problem: there is no "vertical" normal line to this "horizontal parabola"; all of the solutions come directly from the equation as it was found.]
[A property of the normal line through an external point perhaps worth mentioning is that we have incidentally found that $ \ (-32 , -10 ) \ $ is the closest point on the parabola to $ \ (X, Y) $ $ = \ ( -27, -50 ) . $ In turn, it is the closest point on the parabola to any other point on the same side of the normal line. I have said a bit about this topic here.]
As for the problem statement above, it is not entirely accurate, as the point $ \ (X, Y) \ = \ ( - \frac{9}{2}, -5 ) $ does in fact lie on the parabola. However, this does not materially affect our calculations, and already "tips us off" as to what one of the solutions is (since a point on the curve automatically is on its normal line). The cubic polynomial here is $ \ u^3 \ - \ 7 u \ + \ 6 \ = \ 0 \ , $ one of its roots again turns out to be $ \ u = 2 \ , $ and its factorization is $ \ (u + 3) \ (u - 1) \ (u - 2) \ = \ 0 \ . $ [OK, I happened to start off using 2 again: 1 is a more obvious root...] The solution points are therefore $ \ (- \frac{9}{2} , -5 ) \ $ , $ \ (- \frac{1}{2} , -1 ) \ , \ \text{and} \ (-2 , 0 ) \ . $
I'm afraid I can't say that there is much of what I've done here that can be commended as a general method; things work out as tidily as they have largely because of the choices for the "external points".
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.