<s>[INST] Finding the equation of the Parabola The parabola $y = x^2 + bx + c$ has the following properties:

*

*The point on the parabola closest to $(12,3)$ is the intersection with the $y$ axis of the parabola.

*The parabola passes through $(-5,0).$
How can I find $(b, c)$?

Here is my attempt:
The point $(0, c)$ is the intersection with the $y$ axis of the parabola. The distance from $(12, 3)$ is $\sqrt{144 + (c-3)^2}$, and we have the equation $-5b +25+c =0$.
But if we don't know vertex of the parabola how do we find $b$ & $c$?
 [/INST] 

The parabola $y = x^2 + bx + c$

*

*The point on the parabola closest to $(12,3)$ is the intersection of the $y$ axis and the parabola


and we have the equation $-5b +25+c =0$.


*

*Let $P$ be the parabola's $y$-intercept $(0,c)$ and $Q$ be $(12,3).$
Then the parabola's tangent at $P$ is perpendicular to $PQ,$ and,
since $P$ and $Q$ have different $x$-coordinates, $PQ$ is not
vertical. So,
$$(2x+b)\left(\frac{c-3}{0-12}\right)=-1\quad\text{and}\quad
x=0\\b(c-3)=12.$$ And as $c=5b-25,$
$$\\5b^2-28b-12=0\\b=6\quad\text{or}\quad
-\frac25\\c=5\quad\text{or}\quad -27.$$


*If $(b,c)=\left(-\frac25,-27\right),$ then the parabola is
$\displaystyle y=x^2-\frac25x-27,$ and its points whose tangents are
perpendicular to $PQ$ are given by
$$\left(2x-\frac25\right)\left(\frac{x^2-\frac25x-27-3}{x-12}\right)=-1\\x=-5.13\quad\text{or}\quad
0 \quad\text{or}\quad 5.73,$$ and the corresponding distances from
$(12,3)$ are approximately $17,32$ and $7.$ We eliminate this
case since here the point on the parabola closest to $(12,3)$ is
not the $y$-intercept.
SIMPLER ALTERNATIVE (suggested by Oscar Lanzi below): If $b=-\frac25,$ then the parabola's axis $x={-}\frac b2$ is $x=\frac15,$ and $P$ and $Q$ lie on opposite sides of it, which means that some point on the parabola is closer than $P$ to $Q;$ so, we eliminate this case.


*Hence, the required
parabola must be
$y=x^2+6x+5.$
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