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

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

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1 Answer

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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.
answered Jun 14, 2013 by sreemathi.v
edited Jun 23, 2013 by sreemathi.v
 

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