What is the equation in standard form of the parabola with a focus at (14,5)...

Explanation:
Focus is at (14,5) and directrix is y=-15. Vertex is at midway
between focus and directrix. Therefore vertex is at
(14, ${\displaystyle \frac{5-15}{2}}$) or (14, -5). The vertex form of equation of
parabola is ${\displaystyle y=a{(x-h)}^2+k}$; (h.k); being vertex. Here
h=14 and k=-5 So the equation of parabola is
${\displaystyle y=a{(x-14)}^2-5}$. Distance of vertex from directrix is
d=15-5=10, we know ${\displaystyle d=\frac{1}{4\left|a\right|}\therefore \left|a\right|=\frac{1}{4d}}$ or
${\displaystyle \left|a\right|=\frac{1}{4\cdot 10}=\frac{1}{40}}$. Here the directrix is below
the vertex , so parabola opens upward and a is positive.
${\displaystyle \therefore a=\frac{1}{40}}$ Hence the equation of parabola is
${\displaystyle y=\frac{1}{40}{(x-14)}^2-5}$

Explanation:
the standard form of a parabola in translated form is.
${\displaystyle {(x-h)}^2=4p(y-k)}$
where (h,k) are the coordinates of the vertex
and p is the distance from the vertex to the focus
since the directrix is below the focus then the curve
opens upwards
coordinates of vertex ${\displaystyle =(14,\frac{5-15}{2})=(14,-5)}$
and p=5-(-5)=10
${\displaystyle \Rightarrow {(x-14)}^2=40(y+5)\leftarrow }$ equation of parabola