Finding the focus of a parabola given an equation

[BlueQuark]

Homework Statement

Alright, so the equation of a parabola is y = 1/4p*x^2, P being either an x or y value, and the other x or y being zero. Let's say that x^2 = 16y. If you divide both sides by 16, you get y = x^2/16, which can be simplified to y = 1/16*x^2. This is in the format of a parabola, so finding p is simple. 16 is the product of 4p, so 4p=16. Divide both sides by 4 and p=4. So the focus is at (0,4).

This seems simple to me, until you get equations like this: 3x^2 + 4y = 0. I can't seem to get this in the form of a parabola. I got -3/4 *x^2=y. This wouldn't help me find the focus, due to the fact that it's not in that form. How exactly would you go about finding the focus in cases like this? Thanks for any help!

 

[SammyS]

Homework Statement

Alright, so the equation of a parabola is y = 1/4p*x^2, P being either an x or y value, and the other x or y being zero. Let's say that x^2 = 16y. If you divide both sides by 16, you get y = x^2/16, which can be simplified to y = 1/16*x^2. This is in the format of a parabola, so finding p is simple. 16 is the product of 4p, so 4p=16. Divide both sides by 4 and p=4. So the focus is at (0,4).

This seems simple to me, until you get equations like this: 3x^2 + 4y = 0. I can't seem to get this in the form of a parabola. I got -3/4 *x^2=y. This wouldn't help me find the focus, due to the fact that it's not in that form. How exactly would you go about finding the focus in cases like this? Thanks for any help!

Hello BlueQuark. Welcome to PF !

You need parentheses around the 4p since both are in the denominator. You are using LaTeX, so it is easy to use "\frac" to write : $\ y = \frac{1}{4p}x^2\$.

Writing the equation of the parabola as $\ 4py = x^2\$ may prove to be even handier for solving your problem.

 

[member 587159]

4y + 3x^2 = 0
<=> y = -3/4 x^2
<=> y = -1/(4/3)x^2

I don't know whether this has answered your question.

 

[HallsofIvy]

Homework Statement

Alright, so the equation of a parabola is y = 1/4p*x^2

As SammyS said, that should be y= 1/(4p)x^2 or y= x^2/(4p)

, P being either an x or y value, and the other x or y being zero.

I really don't know what that means. What does it mean to say that p is "either an x or y value". What do you mean by an "x value" or a "y value"? And what "other" x or y do you mean? p is, of course, the x coordinate of the focus- in this situation, the focus is at (0, p). Is that what you mean?

Let's say that x^2 = 16y. If you divide both sides by 16, you get y = x^2/16, which can be simplified to y = 1/16*x^2. This is in the format of a parabola, so finding p is simple. 16 is the product of 4p, so 4p=16. Divide both sides by 4 and p=4. So the focus is at (0,4).

This seems simple to me, until you get equations like this: 3x^2 + 4y = 0. I can't seem to get this in the form of a parabola. I got -3/4 *x^2=y. This wouldn't help me find the focus, due to the fact that it's not in that form. How exactly would you go about finding the focus in cases like this? Thanks for any help!

Surely you know that -3/4= \frac{1}{-\frac{4}{3}}? p= -\frac{4}{3} and the focus is at \left(0, -\frac{4}{3}\right).

 

[member 587159]

As SammyS said, that should be y= 1/(4p)x^2 or y= x^2/(4p) I really don't know what that means. What does it mean to say that p is "either an x or y value". What do you mean by an "x value" or a "y value"? And what "other" x or y do you mean? p is, of course, the x coordinate of the focus- in this situation, the focus is at (0, p). Is that what you mean?Surely you know that -3/4= \frac{1}{-\frac{4}{3}}? p= -\frac{4}{3} and the focus is at \left(0, -\frac{4}{3}\right).

Yes, -3/4= \frac{1}{-\frac{4}{3}}

And, y = -x^2/(4p) is a valid equation for a parabola.
P = -1/3, not -4/3 so the focus is at (0, -1/3)