Question:
Find the coordinates of the focus and the vertex, the equations of the directrix and the axis, and length of the latus rectum of the parabola :
$3 x^{2}=-16 y$
Solution:
Given equation :
$3 x^{2}=-16 y$
$x^{2}=-\frac{16}{3} y$
Comparing the given equation with parabola having an equation,
$x^{2}=4 a y$
$4 a=\frac{16}{3}$
$a=\frac{4}{3}$
Focus: $F(0,-a)=F\left(0,-\frac{4}{3}\right)$
Vertex: $A(0,0)=A(0,0)$
Equation of the directrix: $y-a=0$
$y-\frac{4}{3}=0$
$y=\frac{4}{3}$
Lenth of latusrectum :
$4 a=\frac{16}{3}$
