Question

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

Answer 1

@Hero

Answer 2

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

Answer 3

working it meow

Answer 4

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

Answer 5

@Hero

Answer 6

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.

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

Then took the square root of both sides

Answer 13

That still doesn't look right.

Answer 14

You must have made another mistake.

Answer 15

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

Answer 16

left that out by accident

Answer 17

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

Answer 18

Does that look right?

Answer 19

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.

Answer 20

I recommend starting over but posting each step this time

Answer 21

Alright one sec

Answer 22

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

Answer 23

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

Answer 24

Do you see any problems yet? @hero

Answer 25

Yes, I do actually. Double check your work.

Answer 26

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

Answer 27

Is that correct?

Answer 28

oh Sh*t

Answer 29

Can't believe I didn't see that earlier

Answer 30

What didn't you see earlier?

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

Ok. So what you had was correct

Answer 36

The parabola is horizontal instead of vertical.

Answer 37

Oh yeah XD Derp

Answer 38

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

Answer 39

Different type of question

Answer 40

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

Answer 41

ok thanks @hero