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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

Best Answer 

 #1
avatar+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:

 

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

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

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

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

And that is your equation.

Mathhemathh  Oct 23, 2017
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3+0 Answers

 #1
avatar+302 
+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
avatar+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

 

 

cool cool cool

CPhill  Oct 23, 2017
 #3
avatar+78604 
+1

 

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

 

Kudos.....!!!

 

 

cool cool cool

CPhill  Oct 23, 2017

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