Find the coordinates of the focus and the vertex, the equations of the

Solution:

Given equation :

$5 y^{2}=-16 x$

$y^{2}=-\frac{16}{5} x$

Comparing the given equation with parabola having an equation,

$y^{2}=-4 a x$

$\cdot 4 a=\frac{16}{5}$

$\cdot a=\frac{4}{5}$

Focus: $F(-a, 0)$

$=F\left(-\frac{4}{5}, 0\right)$

Vertex :

$\mathrm{A}(0,0)=\mathrm{A}(0,0)$

Equation of the directrix :

$x-a=0$

$x-\frac{4}{5}=0$

$x=\frac{4}{5}$

Lenth of latusrectum : $4 a=\frac{16}{5}$