# Thread: Find equation of the parabola

1. ## Find equation of the parabola

If the axis is parallel to x axis, Latus Rectum = 1 and passing through (3,1) and (-5,5)

Help please. No idea. The axis parallel to x axis made it confusing.

2. Hello reiward
Originally Posted by reiward
If the axis is parallel to x axis, Latus Rectum = 1 and passing through (3,1) and (-5,5)

Help please. No idea. The axis parallel to x axis made it confusing.
The 'standard' parabola with axis parallel to the $x$-axis has equation
$y^2 = 4ax$
and if we move the vertex from the origin to the point $(h,k)$, this equation becomes:
$(y-k)^2 = 4a(x-h)$
Now the latus rectum has length $4a$. So here $4a = 1$, and the equation now becomes:
$(y-k)^2 = (x-h)$
Plug in the values of $x$ and $y$ at the two points you're given:
$\left\{\begin{array}{l}
(1-k)^2=3-h\\
(5-k)^2=-5-h\\
\end{array}\right.$

$\Rightarrow\left\{\begin{array}{l}
k=4\\
h=-6\\
\end{array}\right.$

So the equation is:
$(y-4)^2 = x+6$
which can be re-written as:
$x=y^2-8y+10$
if you like.