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Mathematics 8 Online
OpenStudy (anonymous):

What is the directrix of the parabola defined by (x + 3)2 = -20(y − 1)? A) x = -4 B) y = -4 C) x = 6 D) y = 6

OpenStudy (igreen):

@Haley223 Help this person! They're doing the exact same thing you're doing!

OpenStudy (anonymous):

im on plato

OpenStudy (imstuck):

The standard equation for a parabola is \[(x-h)^{2}=4p(y-k)\]To find the directrix we need to remember that it is exactly the same number of unites from the vertex as the focus is, but in the opposite direction. Do you know from looking at your equation if the directrix is an x = line or a y = line?

OpenStudy (anonymous):

no ive never learned this ._.

OpenStudy (imstuck):

What I'm asking is if you know if the line is "x = " something or "y = " something.

OpenStudy (anonymous):

no ................ i dont ;-;

OpenStudy (imstuck):

You never learned this? Ok...here's the deal, then.

OpenStudy (anonymous):

...

OpenStudy (imstuck):

This is what a parabola looks like, centered at the origin, with the focus and the directrix.|dw:1403010745800:dw|That is an example. The focus and the directrix are an equal distance away from the vertex.

OpenStudy (imstuck):

In order to find out the location of your focal point, and then from that directrix line, you need to solve for p in your equation\[(x-h)^{2}=4p(y-k)\]This is a parabola that opens upward like a cup, because it is an x^2 parabola and there isn't a negative sign with the x^2. Get that part?

OpenStudy (imstuck):

Sorry, it is an upside down opening parabola since there is a negative with the y. Now if we solve our equation for p, we can find out what the focus is and then the directirx line. Since this is an x^2 parabola, the directrix a line of the form "y = " something. We need to find that something. Finding the focus will help us. If 4p=y from our equation, we have 4p = -20. Solving for p we get p = -5. This would be the focus if the parabola were centered at the origin. It would look like this|dw:1403011214543:dw|because the graph of a parabola will always wrap itself around the focal point.

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