Question

Find the vertex, axis, directrix, focus, latus rectum and the tangent at vertex for the parabola $9y^2-16x-12y-57=0$.

Answer 1

$9y^2-16x-12y-57=0$

So ${(y-\frac{2}{3})}^2=\frac{16}{9}(x+\frac{61}{16})$

i.e; $y^2=4ax$.

Vertex is $(\frac{-61}{16},\frac{2}{3})$.

This axis is $y-\frac{2}{3}=0$

$y=\frac{2}{3}$

directrix is $:x+a=0$

$x+\frac{61}{16}+\frac{4}{9}=0$

$x=-613/144$.

Focus is $x=a$ & $y=0$

$x+61/16=4/9$ and

$y-2/3=0$.

$(-485/144,2/3)$ is focus

Length of latus rectum

$=4A=16/9$

Tangent at vertex is

$X=0$

$X=\frac{-61}{16}$.