An equation of the form $x ^ { 2 } = 4 p y$ is given. Identify the focus and the endpoints of the latus rectum.
-$$x ^ { 2 } = - 12 y$$
A) focus: $( 0 , - 3 )$; endpoints of latus rectum: $( - 12 , - 3 ) ; ( 12 , - 3 )$
B) focus: $( 0,3 )$; endpoints of latus rectum: $( - 12,3 ) ; ( 12,3 )$
C) focus: $( 0 , - 3 )$; endpoints of latus rectum: $( - 6 , - 3 ) ; ( 6 , - 3 )$
D) focus: $( 0,3 )$; endpoints of latus rectum: $( - 6,3 ) ; ( 6,3 )$

Question

Quizplus · Accepted Answer

The given equation is $x^2 = -12y$, which can be rewritten as $x^2 = 4(-3)y$, indicating that $p = -3$. The focus of a parabola in the form $x^2 = 4py$ is $(0, p)$, so here it is $(0, -3)$. The latus rectum is $4|p|$ units long and centered at the focus, so its endpoints are $(-2|p|, p)$ and $(2|p|, p)$, which for $p = -3$ gives $(-6, -3)$ and $(6, -3)$.

An Equation of the Form x2=4pyx ^ { 2 } = 4 p yx2=4py

An Equation of the Form $x ^ { 2 } = 4 p y$