Question

Help plz!!
Find the standard form of the equation of the parabola with a focus at (3, 0) and a directrix at x = -3.

Answer 1

y = one divided by twelvex2

-12y = x2

x = one divided by twelvey2

y2 = 6x

Answer 2

Those are the possible answers and I think its the third one
I just need clarification

Answer 3

"I think its the" - Nope. Never good enough. Prove it!

Definition of Parabola. Set of points equidistant from a given point (focus) and a given line (directrix).

Given Point: (3,0)
Given Line: x = -3

What is the distance of ANY point, (x,y) from (3,0)?
What is the distance of ANY point, (x,y) from x = -3?

Answer 4

3 and -3 I'm assuming the (x,y) is at the origin?
(totally clueless)

Answer 5

I said ANY point, not the Origin.

ANY point, (x,y) is this far from (3,0), \(\sqrt{(x-3)^{2}+(y-0)^{2}}\). The distance formula. Are we ringing any bells?

Answer 6

Oh! I know the formula but to use it for this wasn't in the lesson.

Answer 7

ok so from that equation I cant really get an answer so do I plug in values for the x and y?

Answer 8

If you know it, it's for this lesson. Remember EVERYTHING.

Okay, what is the distance of ANY point, (x,y) from x = -3? Try a few points and see.

Answer 9

Well I plugged in (1,1) into the equation and got Sqrt 5
Did it again from (2,7) and got 5Sqrt2

Answer 10

No, we're no longer worried about the distance from (3,0). We have that, already. It's that lovely square root from the distance formula. NOW, we need to know the distance from x = -3.

Answer 11

So what formula asre yu saying I should be using?

Answer 12

Are you saying to use slope or the same distance formula?

Answer 13

If we are going from the focus then the distance would be 6

Answer 14

Geometry. Draw a perpendicular line from x = -3 through the desired point. Measure the distance.

The point (0,4) is a distance of 3 from x = -3.

The point (1,3) is a distance of 4 from x = -3.

The point (-2,2) is a distance of 1 from x = -3.

You have to see it!

Answer 15

I see that. Sorry didnt understand you meant from random points again.

Answer 16

They are NOT random points. We are talking about EVERY CONCEIVABLE point.

Answer 17

Given a point, I don't care which one, how far is it from the focus?
Given that same point, whichever one we picked in the previous quiz, how far is it from x = -3?

Answer 18

Ok so from (0,4) the distance would be 3 and from it would be 6 from the focus

Answer 19

*and it would be 6 from the focus

Answer 20

We don't care about (0,4). That is just one sample point from infinitely many points. We have to get them ALL.

All points are this far from (3,0): \(\sqrt{(x-3)^{2}+(y)^{2}}\)

All points are this far from x = -3: \(|x-(-3)| = |x+3|\)

You must see these two things. You can pick ANY point and calculate those two distances. However, we don't have a parabola until we do this:

\(\sqrt{(x-3)^{2}+y^{2}} = |x+3|\)

Now THAT is a parabola. Your task is to put it in standard form.

Answer 21

oh ok!! Yay! So for problems like these in the future do I just follow the set up that we just went through?

Answer 22

I hope not. We should go directly to the solution, rather than walking around the block a few times.

I hope not. This is the focus-directrix DEFINITION of a parabola. It is useful mostly for proving existence. Definitions are often cumbersome therefore, not particularly helpful for more common use.

I hope not. You should learn a couple other methods along the way. Keep your eyes and ears open and understand what you are doing. Don't try to memorize some process.

Answer 23

Okay, thank you so much!
I got x=1/12y^2 btw :)

Answer 24

Is that right?

Answer 25

Yep! :)

Answer 26

Perfect.