Answer
See below
Work Step by Step
Given $x^2-16x-8y+80=0$
We can see that $a=1\\b=0\\c=0$
We will find the discriminant of the given equation $=b^2-4ac\\=0^2-4(1)(0)\\=0$
The conic is a parabola.
To graph the ellipse, first complete the square:
$x^2-16x-8y+80=0\\x^2-16x=8y-80\\(x-8)^2=8y-16\\(x-8)^2=8(y-2)$
From the equation, you can see that the vertex is at $(8,2)$.
