Question

What will be the end points of latus rectum of translated parabola $x=y^2+2y+5$?

Answer 1

Given, the equation of translated parabola as $x=y^2+2y+5.$
On rearranging the equation we get,
$\Rightarrow (x-4)=(y+1{)}^2$
$\Rightarrow (y+1{)}^2=4\times \frac{1}{4}(x-4)$
On comparing with equation of translated parabola we get the vertex of parabola is $(4,-1)$ and $a=\frac{1}{4}.$
Now, the coordinates of end point of latus rectum is given by $(h+a,k+2a),(h+a,k-2a)$ i.e. $(\frac{17}{4},-\frac{1}{2}),(\frac{17}{4},-\frac{3}{2}).$