Browse Questions

# Find the equation of parabola if : Vertex $(0 , 0 );$ focus : $(0 , -4 )$.

This is the third part of the multi-part question Q1

Toolbox:
• Standard parabolas :
• (i)$y^2=4ax$
• In this case the axis of symmetry is $x$-axis.
• The focus is $F(a,0)$ and the equation of the directrix is $x=-a$
• (ii)$y^2=-4ax$
• In this case the focus is $F(-a,0)$ and the equation of directrix is $x=a$
• (iii)$x^2=4ay$
• In this case the axis of symmetry is $y$-axis.
• The focus is $F(0,a)$ and the equation of directrix is $y=-a$
• (iv)$x^2=-4ay$
• In this case the focus is $F(0,-a)$ and the equation of directrix is $y=a$
Step 1:
Vertex $V(0,0)$,focus $F(0,-4)$
$VF=a=4$units.
The vertex is at the origin.
The axis is along $VF$ is y-axis($x=0)$.
Step 2:
$F$ lies on the -ve side and so the parabola has an equation of the form $x^2=-4ay$
$a=4$
$\Rightarrow x^2=-4\times 4y$
$x^2=-16y$
This is the required equation.
edited Jun 23, 2013