Find the focus and directrix of the parabola

**htk** · June 8th 2010, 03:41 PM

Find the focus, directrix and axis of the parabola with equation x= 2y^2+3?
x-3= 2y^2

x-3 = y^2 -------------> Is it how you solve it?

Thank you for helping me!

**earboth** · June 8th 2010, 10:15 PM

Originally Posted by htk

Find the focus, directrix and axis of the parabola with equation x= 2y^2+3?
x-3= 2y^2

x-3 = y^2 -------------> Is it how you solve it?

Thank you for helping me!

1. This is a parabola opening to the right. The general equation of such a parabola is:

$(y-h)^2=4p(x-k)$ where V(k, h) is the vertex, F(k+p, h) is the focus and d: x = k-p is the equation of the directrix.

2. Transform the given equation into the general form:

$x= 2y^2+3~\implies~(y-0)^2=\frac12(x-3)~\implies~(y-0)^2=4 \cdot \frac18(x-3)$

3. Determine V, F and d.

**htk** · June 9th 2010, 03:53 AM

So, V (3,0) F=(25/8, 0) D x= 23/8 Are these correct? Thanks again!

4p= 1/2
p= 1/8

**earboth** · June 9th 2010, 06:56 AM

Originally Posted by htk

So, V (3,0) F=(25/8, 0) D x= 23/8 Are these correct? Thanks again!

4p= 1/2
p= 1/8

Perfect!

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