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.
Grandad