Find a point on a parabola that's closest to another point.
Find the point on the parabola $3x^2+4x-8$ that is closest to the point $(-2,-3)$.
My plan for this problem was to use the distance formula and then that the derivative to get my answer. I'm having a little trouble along the way.
$$ d = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}.$$
$\endgroup$ 12 Answers
$\begingroup$$$ d = \sqrt{(x - (-2))^2 + ((3x^2+4x-8) - (-3))^2} = \sqrt{\left(x+2\right) ^2 + \left(3x^2+4x-5 \right)^2}$$
Instead of minimizing $d$, lets minimize $d^2$. It should be noted that the minimum of $d$ and $d^2$ are exactly the same as $d \geq 0$. This gets rid of the square root, thereby simplifying the problem.
$$d^2 = (x+2)^2 + (3x^2+4x-5)^2.$$
Now, we can use calculus to minimize $d^2$,
$$(d^2)' = 2(x+2) + 2(3x^2+4x-5)(6x+4) = 36x^3 + 72x^2 - 26x - 36$$
This has roots at $x \approx -2.118, -.631, .749$. We know that one of these must be the absolute minimum, so evaluate $d^2$ for each $x$-value and whichever is smallest is the $x$-value of the point on the parabola nearest to the given point. You should then plug that $x$-value into the equation for the parabola to determine the $y$-value where this occurs. I got,
$\endgroup$ $\begingroup$Ans: $ (-2.118, -3.014) $
Suppose the closest point is at $p=(x_0,y_0)$, and set $q=(-2,-3)$. Then the tangent to the parabola at $p$ is perpendicular to $\ell$, the line through $p,q$.
Since $y'=6x+4$, the slope of the tangent is $6x_0+4$, so the slope of $\ell$ is $-\frac{1}{6x_0+4}$. Since it passes through $q$, we see the equation for $\ell$ is $$y+3 = -\frac{x+2}{6x_0+4}$$
Combining with $y=3x^2+4x-8$ shows that $x_0$ is a root of $$ (6x+4)(3x^2+4x-5)+(x+2)=18x^3+36x^2-13x-18=0 $$ and we end up numerically computing the roots from here.
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