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September 21, 2014

September 21, 2014

Posted by **Vincent** on Tuesday, February 3, 2009 at 9:41am.

Passes through the point (4,6); horizontal axis.

I know it has to be y^2=4px. But I don't know what to do with (4,6). Please show step by step. Thanks.

- Algebra2/Trig -
**Damon**, Tuesday, February 3, 2009 at 9:55amHorizontal axis means of form:

(y-k)^2 = 4p(x-h)

if the vertex is at the origin (h,k) = (0,0)

y^2 = 4 p x

36 = 4 p (4)

p = (9/4)

so

y^2 = 9 x

- Algebra2/Trig -
**Vincent**, Tuesday, February 3, 2009 at 10:06amOh, I see. Solve for P first then substitute it in. I guess I forgot that concept. Thanks!

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