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# How To Find Parabola?

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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
+302
+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:

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

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

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

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

Mathhemathh  Oct 23, 2017
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#1
+302
+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:

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

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

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

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

Mathhemathh  Oct 23, 2017
#2
+78604
+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
#3
+78604
+1

WOW, Mathhemathh....I've not seen that method before.....but....I like it  !!!!

Kudos.....!!!

CPhill  Oct 23, 2017

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