Question

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

Answer 1

Given equation of the parabola

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

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

$9(y^2-\frac{4}{3}y)=16x+57$

$[y^2-2\left(\frac{2}{3}\right)y+\frac{4}{9}]-\frac{4}{9}=\frac{16x}{9}+\frac{57}{9}\Rightarrow {(y-\frac{2}{3})}^2=\frac{16x}{9}+\frac{57}{9}+\frac{4}{9}$

$\Rightarrow {(y-\frac{2}{3})}^2=\frac{16}{9}[x+\frac{61}{16}]....\left(1\right)$

The above equation is of the form $y^2=4Ax......\left(2\right)$

${(y-\frac{2}{3})}^2=\frac{16}{9}[x-\left(\frac{-61}{16}\right)]$

Therefore $h=\frac{-61}{16},k=\frac{2}{3}$

$\therefore$ Vertex $A(\frac{-61}{16},\frac{2}{3})$

On comparing equations (1) and (2) $A=\frac{4}{9}$

Focus $=(h+a,k)=(\frac{-61}{16}+\frac{4}{9},\frac{2}{3})=(\frac{-485}{144},\frac{2}{3})$

Equation of the parabola is $y=0$

$\therefore y-\frac{2}{3}=0,y=\frac{2}{3}$

Directive equation of directive is $x+a=0$

$x+\frac{61}{16}=\frac{-4}{9}\Rightarrow x=\frac{-613}{144}$

Length of latus rectum $=4A=4\left(\frac{4}{9}\right)=\frac{16}{9}$