Question

Find the vertex, axis focus, directrix and latus rectum of the parabola $y^2+4x+4y-3=0$.

Answer 1

$Y^2+4y+4x-3=0$

$(y+2{)}^2-4+4x-3=0$

$(y+2{)}^2=-4(x-\frac{7}{4})$

Let $Y=y+2,X=x-\frac{7}{4}$

Then we have

$Y^2=4x$

Comparing the gives equation with $Y^2=-4ax$

$4a=4\Rightarrow a=1$

$\therefore$ vertex $=(X=0,Y=0)=(x=\frac{7}{4},y=-2)$

Focus $=(X=-a=-1,Y=0)(x-\frac{7}{4}=-1,y+2=0)=(x=\frac{3}{4},y=-2)$

Equation of directrix

$X=a=1$

i.e $x-\frac{7}{4}=1\Rightarrow x=\frac{11}{4}$

Axis $=Y=0$

i.e $y+2=0\Rightarrow y=-2$

length of the latus rectum $=4a=4$ units