Question

If a parabola's focus is at (2,4) and directrix is at y=-2, what is the vertex form of the equation representing this parabola?

Answer 1

Answer:

$y=\frac{(x+2)^2}{14}+\frac{3}{2}$

Step-by-step explanation:

When we have a focus and the directix of a parabols, we find the vertex form of the equation by using the distance formula. Our focus is (2,4) with directix of y=-2. We take any point on the parabola (x,y)

$\sqrt{(x-2)^2+(y+4)^2}$ is the distance from the focus to some point (x,y) on the parabola. We know that this distance will be equal to the distance from  the (x,y) point to the directix at y=-2. We write $\sqrt{(y+2)^2}$. We now set them equal and simplify to vertex form.

$\sqrt{(y+2)^2}=\sqrt{(x+2)^2+(y-5)^2}$

$y^2+4y+4=(x+2)^2+y^2-10y+25$

After expanding the y terms, we move across the equation sign using the inverse to reduce to smallest terms.

$y^2-y^2+4y+4=(x+2)^2+y^2-y^2-10y+25\\4y+4=(x+2)^2+-10y+25\\4y+10y+4=(x+2)^2+-10y+10y+25\\14y+4-4=(x+2)^2+25-4\\14y=(x+2)^2+21\\y=\frac{(x+2)^2+21}{14} \\y=\frac{(x+2)^2}{14}+\frac{3}{2}$