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 $b$ & $c$?
 A: 

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
$y=x^2+6x+5.$
A: $ y = x^2 + bx + c $
$ y' = 2 x + b $
At $(0,c)$ we have $y' = b $, and thus
$ b \bigg( \dfrac{c - 3}{0 - 12} \bigg) = -1 $
which simplifies to $ b (c - 3) = 12 $
We also have $(-5, 0)$ on the parabola, then  $0 = 25 - 5 b + c $
Solving the above two equations yields two solutions
Case I:  $b = -0.4, c = -27 $
Case II: $ b = 5 , c = 6 $
We need to check if the distance at the $y-intercept$ is indeed the shortest distance.  At the critical points, we have
$ (2 x + b ) \bigg( \dfrac{x^2 + bx + c - 3 }{ x - 12 } \bigg) = -1 $
So that
$ 2 x^3 + 3 b x^2 + x ( 1 + 2  (c - 3) + b^2) + b ( c - 3) - 12 = 0 $
In case I, this gives three critical points, listed with their respective distances from $(12, 3)$
$A (-5.13231, 1.393538) , 17.20746 $
$B (0, -27) , 32.31099 $
$ C (5.732311, 3.566462), 6.293235 $
Clearly the $y$-intercept, which is point $B$ is not the closest point to $(12,3)$.  Hence Case I is extraneous.
Case II leads to one critical point only which is $(0, 5)$ with a distance of $12.16553$.
Hence our parabola is
$ y =x^2 + 6 x + 5 $
