how to find focal radius in parabola?

Question

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.

Thanks.

Answer 1

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 $p$.

Answer 2

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.

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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 $f$.

Answer 3

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