Question

Find the vertex, focus, axis, latusrectum of the parabolas reducible to the standard form. $y^2-8y-x+19=0$

Answer 1

Given that,

$y^2-8y-x+19=0$

Rearranging the above equation of parabola to separate x and y, we get:

$y^2-8y+16=x-3$

$(y-4{)}^2=x-3$

Now, equating the above equation with the standard form of parabola, $(y-k{)}^2=4a(x-h),$ we get:

$h=3,k=4,a=1/4$ where $(h,k)$ is the coordinate of vertex and a is the focal length.

The axis of the parabola is x axis because the parabola is symmetric about $x$ axis. The parabola has its face opened toward positive $x$-axis.

So,$vertex=(3,4)$

$axis=x-axis$

$focus=(3+1/4,4)=(13/4,4)$

$directrix,x=3-1/4=11/4$

$Latusrectum,LR=4a=4\times \frac{1}{4}=1$