A parabola has a focus of F(−1,5) and a directrix of y=6. What is the equation of the parabola?

Guest Oct 23, 2017

#1**+2 **

Using distance formula, the distance of the parabola from its focus is \(\sqrt{(x+1)^2+(y-5)^2}\) and the distance from the directrix is \(\sqrt{(y-6)^2}\). On aparabola, these distances are always equal, so:

\(y-6=\sqrt{(x+1)^2+(y-5)^2}\)

\(y^2-12y+36=(x+1)^2+y^2-10y+25\)

\(-2y+11=(x+1)^2\)

\(y=-{(x+1)^2-11\over2}\)

And that is your equation.

Mathhemathh
Oct 23, 2017

#1**+2 **

Best Answer

Using distance formula, the distance of the parabola from its focus is \(\sqrt{(x+1)^2+(y-5)^2}\) and the distance from the directrix is \(\sqrt{(y-6)^2}\). On aparabola, these distances are always equal, so:

\(y-6=\sqrt{(x+1)^2+(y-5)^2}\)

\(y^2-12y+36=(x+1)^2+y^2-10y+25\)

\(-2y+11=(x+1)^2\)

\(y=-{(x+1)^2-11\over2}\)

And that is your equation.

Mathhemathh
Oct 23, 2017

#2**+1 **

This parabola will turn downward since the directrix is above the foucus

The vertex will be at ( -1, [ sum of directix and y coordinate of the vertex]/2 ) =

(-1, [ 6 + 5 ] / 2 ) = (-1, 11/2)

And the distance between the focus and the vertex = p = -.5 = -1/2

And the equation is

4p ( y - 11/2) = ( x + 1)^2

4 (-1/2) ( y - 11/2) = (x + 1)^2

-2 ( y - 11/2) = ( x + 1)^2

y - 11/2 = (-1/2)( + 1)^2

y = (-1/2)(x + 1)^2 + 11/2

Here's the graph : https://www.desmos.com/calculator/zfnw68cp1f

CPhill
Oct 23, 2017