Find the standard form of the equation of the parabola with a focus at (-8, 0) and a directrix at x = 8.

- anonymous

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- anonymous

@Hero

- Hero

Okay, go ahead and solve it using the formula I gave you previously. This time, the point for the directrix will be (8,y).

- anonymous

working it meow

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## More answers

- anonymous

I got \[y=x-8\sqrt{2}\]

- anonymous

@Hero

- Hero

You should show the complete work you did on it. From what I observe, your final answer doesn't seem to have the right form.

- anonymous

Im skipping the first step cause It takes so long to write
\[\large x^2 + 64 +y^2 = x^2 - 64 +x^2 - y^2\]

- anonymous

@Hero \[\large 64 = x^2 - y^2 - 64\]
I feel like this is a formula of some sort...

- Hero

Skipping steps doesn't sound like such a good idea.

- anonymous

I didn't skip any but the first where you plug in the values

- anonymous

next i did this: \[\large y^2 = x^2 - 128\]

- anonymous

Then took the square root of both sides

- Hero

That still doesn't look right.

- Hero

You must have made another mistake.

- Hero

BTW, \(y^2 + y^2 = 2y^2\)

- anonymous

left that out by accident

- anonymous

\[\large 64 = -2y^2 +x^2 -64\]

- anonymous

Does that look right?

- Hero

I wrote out the steps by hand myself. There's a reason why I wanted you to post each step you did just like last time. If you skip steps, you end up confusing yourself. Also, if you never got to the point where you end up using difference of squares, then you may have made yet another mistake.

- Hero

I recommend starting over but posting each step this time

- anonymous

Alright one sec

- anonymous

\[\large (x+8)^2 +y^2 = (x-8)^2 + (x-y)^2\]

- anonymous

\[\large then: x^2 +64 +y^2 = x^2 -64 +x^2 - y^2\]

- anonymous

Do you see any problems yet? @hero

- Hero

Yes, I do actually. Double check your work.

- anonymous

\[\large x^2 + 64 +y^2 = x^2 +64 +x^2 +y^2\]

- anonymous

Is that correct?

- Hero

|dw:1435529904593:dw|

- anonymous

oh Sh*t

- anonymous

Can't believe I didn't see that earlier

- Hero

What didn't you see earlier?

- anonymous

in the last parentheses in the formula I put an x instead of a y

- anonymous

And when I simplify that down it comes out to \[\large y=\sqrt{-32x}\]

- anonymous

unless I did that extremely wrong which I felt like I did

- Hero

Yes, because you should have isolated x instead of y.

- anonymous

Ok. So what you had was correct

- Hero

The parabola is horizontal instead of vertical.

- anonymous

Oh yeah XD Derp

- anonymous

I have some more if you don't mind :)

- anonymous

Different type of question

- Hero

Actually, I'm about to log off soon. Others should be available to help.

- anonymous

ok thanks @hero

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