I'm new to this website and how it works, but I would appreciate it if somebody helped me!
Find the standard form of the equation of the parabola with a focus at (7, 0) and a directrix at x = -7.

- anonymous

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- PhantomCrow

What did you notice when you graphed these two points on a plane?

- anonymous

@Phantomcrow It opens to the right?

- PhantomCrow

Yes! And you know what the standard equation of a parabola that opens to the right is?

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## More answers

- anonymous

@Phantomcrow Can you help me out with that one

- anonymous

It is 3, you're welcome.

- PhantomCrow

Sure.|dw:1444091089825:dw|

- PhantomCrow

Since the vertex is at the origin, there is no need to add any constants for shifts. The graph is simply \[y^2=x\]

- anonymous

@Phantomcrow
By the way the Answer choices are
y= 1/28x^2
x= 1/28y^2
-28y= x^2
y^2=14x
What confuses me is how these tie to the standard formula

- PhantomCrow

D. Despite it having an 'x' term, its vertex is still at the origin.

- PhantomCrow

Standard form is one of the ways to write the graph of a parabola. It is commonly seen as:\[ax^2+bx+c\]

- anonymous

@Phantomcrow Wow thanks, would it be cool if you helped me with one more?

- PhantomCrow

Sure

- anonymous

@Phantomcrow Find the standard form of the equation of the parabola with a focus at (0, 8) and a directrix at y = -8, i'm guessing this is similar or even exact to the last one, so it opens to the left

- PhantomCrow

Actually, it does not open to the left this time. You are given y values for focus and directrix, not x values.

- anonymous

@Phantomcrow So y values open to the right and x values open to the left?

- PhantomCrow

Y values lie across a vertical line, you your parabola will be opening upwards. You can easier see this if you look at the (0.8) and y= -8 on a graph. See that the they are equidistant from a point.

- anonymous

@PhantomCrow you graph (0,8) correct?

- PhantomCrow

Yes. Graph the point (0,8) and y= -8 and you should be able to visually see the the point that is equidistant from them. There is a formula for finding distance but I believe simply looking at the graph for this problem will provide you with the answer.

- anonymous

@PhantomCrow The choices are
y= 1/32x^2
y^2= 8x
y^2=32x
y=1/8x^2
Would it be B?

- PhantomCrow

Oh, I see.

- PhantomCrow

Well, you have two possible answers. A and D. B and C are ruled out because that would be a parabola who's focus is (x,0) and directrix is x=something. We have a focus like (0,y) and y=something in this case so its not y^2=x

- PhantomCrow

Since you need to figure out which is correct between A and D, your best method is to plug in what you know into vertex form (\[(x-h)^2=4p(y-k)\] where (h, k) is the vertex and p is the distance from either the directrix or the focus to the vertex.

- PhantomCrow

We know that the vertex lies at (0,0) because that is the point equidistant from (0,8) and y=-8. P is not difficult to find. P is just 8 because the distance from either the directrix or the focus to the vertex is 8 units (if you need the formula, just ask, but its not necessary).

- PhantomCrow

Plug that into the vertex form:\[(x-h)^2=4p(y-k)\] \[(h,k)=(0,0)\] so \[h=0, k=0\] \[(x-0)^2=4(8)(y-k)\] \[x^2=32y\] \[\frac{ x^2 }{ 32 }=y\]

- PhantomCrow

We want it in terms of y because that's how standard form is written. So we solved for y.

- anonymous

@PhantomCrow Which will result in y=1/32x^2

- PhantomCrow

Yep.

- anonymous

@PhantomCrow Thank you so much for your help have a blessed day!

- PhantomCrow

No problem, you as well.

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