# 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 $\displaystyle x$-axis has equation
$\displaystyle y^2 = 4ax$
and if we move the vertex from the origin to the point $\displaystyle (h,k)$, this equation becomes:
$\displaystyle (y-k)^2 = 4a(x-h)$
Now the latus rectum has length $\displaystyle 4a$. So here $\displaystyle 4a = 1$, and the equation now becomes:
$\displaystyle (y-k)^2 = (x-h)$
Plug in the values of $\displaystyle x$ and $\displaystyle y$ at the two points you're given:
$\displaystyle \left\{\begin{array}{l} (1-k)^2=3-h\\ (5-k)^2=-5-h\\ \end{array}\right.$

$\displaystyle \Rightarrow\left\{\begin{array}{l} k=4\\ h=-6\\ \end{array}\right.$
So the equation is:
$\displaystyle (y-4)^2 = x+6$
which can be re-written as:
$\displaystyle x=y^2-8y+10$
if you like.