Question

What will be the end points $A,B$ of latus rectum of the parabola $(y-1{)}^2=-16(x+3)?$

Answer 1

Given the equation of parabola $(y-1{)}^2=-16(x+3).$
On comparing it with $(y-k{)}^2=4a(x-h)$ we get,
Vertex as $P(-3,1),a=-4$
$\Rightarrow$ Focus will be $S(h+a,k)$ i.e $S(-7,1).$
Now, the coordinates of latus rectum will be $(h+a,k\pm 2a)$ i.e $(-7,9)\&(-7,-7).$
$\therefore$ End points of latus rectum are $A(-7,9),$ $\&B(-7,-7).$