# Focus of a parabola

1. Oct 27, 2009

### Mentallic

1. The problem statement, all variables and given/known data
This problem is from the Australian HSC mathematics extension 2 exam. Q6ciii)

It states:

Find the focus, S, of the parabola $$y^2=r^2+c^2-2cx$$ where r and c are constants.

3. The attempt at a solution
I couldn't figure out how to convert this into the parabola focus form (which, from the top of my head) might be $$(x-x_o)^2=4a(y-y_o)$$ for the focus $$S(x_o,y_o+a)$$

How is this done?

2. Oct 27, 2009

### lanedance

notice the form of x & y are reversed, so you will be looking the equation in the form

$$(y-y_o)^2=4a(x-x_o)$$

with cordinates in the focus changed as well

note that the co-efficient of y is zero, which implies y_o must be zero in the above... see how you go form here

3. Oct 27, 2009

### Staff: Mentor

In your formula, the parabola's vertex is at (x0, y0) and it opens upward if a > 0. The parabola you're working with opens to the left if c > 0, and to the right if c < 0.

Can you put your parabola in the form (y - y)2 = 4a(x - x0)?

4. Oct 27, 2009

### Mentallic

Oh ok so $$r^2+c^2-2cx\equiv 4a(x-x_o)$$

$$LHS=-2c(x-b)=-2cx+2cb$$

therefore $$2cb=r^2+c^2$$

then $$b=\frac{r^2+c^2}{2c}$$

Finally, $$y^2=-2c(x-\frac{r^2+c^2}{2c})$$

So then the focus is $$S(\frac{r^2+c^2}{2c}-\frac{c}{2},0)$$

Is this correct?

edit: simplified, $$S(\frac{r^2}{2c},0)$$

5. Oct 27, 2009

### lanedance

i haven't checked the original focus defintion, but fr0m what you give I get

$$y^2=r^2+c^2-2cx$$

$$(y - 0)^2=4 \frac{1}{4} (r^2+c^2-2cx)$$

$$(y - 0)^2=4 (\frac{-c}{2})(x-\frac{r^2+c^2}{2c})$$

so
$$a = \frac{-c}{2}$$

$$x_0 = \frac{r^2+c^2}{2c}$$

$$y_0 = 0$$

then
$$focus = ((x_0 + a), y_))$$

$$focus = ((\frac{r^2+c^2}{2c} + \frac{-c}{2}), 0)$$

which look the same

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