Question

Derive the equation of the parabola with a focus at (0, –4) and a directrix of y = 4. (2 points)

Answer 1

The vertex of the parabola is between the shortest line connecting its focus and directrix, so between (0, -4) and y = 4 is the point (0, 0), which is the vertex of the parabola. Also, this parabola faces downward.
The general formula for a parabola that faces downward is y = -4cx^2, where c is the distance from the vertex to either to focus or the parabola. Since c = 4, the equation is y = -16x^2, which is choice A.

Answer 2

Equation of parabola with focus at (0,-4) and directrix is y=4 .

As we know parabola is the locus of all the points such that distance from fixed point on the parabola to fixed line directrix is same.

The parabola is opening downwards.

Let any point on parabola is (x,y).

Distance from focus(0,-4) to (x,y) = $\sqrt{(x-0)^{2} +(y+4) ^{2}}=\sqrt{x^{2}+ (y+4)^{2}}$

Distance from (x,y) to directrix, y=4 is =$\left | y-4 \right |$

As these distances are equal.

$\sqrt{x^{2}+ (y+4)^{2}}=\left | y-4 \right |\\{x^{2}+ (y+4)^{2}=(y-4)^{2}$

→x²+y²+8 y +16 = y² - 8 y+16

→ x² = -8 y - 8 y= -16 y   [ Cancelling y² and 16 from L.H.S and R.H.S ]

So , equation of parabola is , x²= - 16 y or f(x)= -x²/16