The parabola with focus at $(4,-3)$ and vertex at $(4,-1)$ is
The parabola with focus at $(4,-3)$ and vertex at $(4,-1)$ is
$\mathrm{x}^2+8 \mathrm{x}+6 \mathrm{y}+22=0$
$x^2-8 x-10 y+6=0$
$x^2-8 x-16 y=0$
$x^2-8 x+8 y+24=0$
Solution
Since points $(4,-3)$ and $(4,-1)$ lie on the line $x=$ 4. So axis of parabola is the line $x=4$. Focus of the parabola lies below the vertex. So the parabola is downwards.
$\therefore \mathrm{a}=\sqrt{(4-4)^2+(-3+1)^2}$ (distance between focus and vertex.
$
=2
$
$\therefore$ Equation of parabola is
$
\begin{aligned}
& (x-4)^2=-4(2)(y+1) \\
& \Rightarrow x^2-8 x+8 y+24=0
\end{aligned}
$