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will we find focal radius in parabol, if our equation is $y^2=12x$. Do I need another variable? I have tried many times but I cannot find this problem.


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It is very rare and unlikely that adding more variables to a problem will make it easier to solve. – rschwieb Jan 15 '13 at 18:54

Hint: When your parabola is written in the form $y=a(x-h)^2+k$ for constants $a,k,h$, the focal length $f$ is related to the constant $a$ by: $a=\frac{1}{4f}$.

Your equation is not in this form, but as a further hint I'll remind you that it's OK to swap the x's and the y's.

Alternatively you can learn "the hard way" by using the definition of the parabola with focus and diretrix. According to the definition of the parabola, the points on the parabola are equidistant from the focus and directrix. If you set up an equation to express the distance between a point $(x_0,y_0)$ on the parabola and call the focus $(f,0)$, you should be able to derive:

$$ \sqrt{(x_0-f)^2+y_0^2}=x_0+f $$

Then you would be able to solve this for $f$.

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We have a parabola with a horizontal axis of symmetry, with vertex at $(h, k)=(0,0)$.

Such a parabola can be written in the form $$(y-k)^2 = 4p(x - h)$$ where $p$ is the distance from the vertex to the focus and the vertex to the directrix.

So, in this case, $$y^2 = 12 x \quad \implies \quad (y-0)^2 = 4\cdot3(x - 0)$$

Now, just identify the value of $p$.

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can I find focal radius with finding the value of p. – Elmi Ahmadov Jan 15 '13 at 19:18
Yes, indeed, Elmi. The focal radius in this case is the length $p$ (which is distance from vertex to the focus, and is also the distance from the vertex to the directix) given by $4p = 12 \implies p = 3$. The focus itself of the parabola is at $(3, 0)$. The vertex of the parabola is at the origin, and it opens toward the right (in the positive direction). – amWhy Jan 15 '13 at 19:22

Focus of parabola $$(y-y_0)^2=2p(x-x_0),y^2=12x,(y-0)^2=2\cdot6(x-0)$$ is $$F(x_0+p/2,y_0)=(0+6/2,0)=(3,0)$$

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