Question

Find the equation of the parabola whose-
Vertex $(0,4)$; Focus $(0,2)$

Answer 1

The coordinates of the vertex is $(0,4)$.

The coordinates of the focus is $\left(0.2\right)$.

It is clear that the vertex and the focus lies on the positive side of the $y$-axis.

Hence the curve is open downwards.

The equation of the form $(x-h{)}^2=-4a(y-k)$

here $h=0,k=4$

$(x-0{)}^2=-4\times 2(y-4)$.

On simplifying we get,

$x^2=-8(y-4)$

or, $x^2=-8y+32$

or, $x^2+8y=32$ is the required equation of the parabola.