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Find the equation of parabola if : Vertex $(1 , 4 );$ focus :$(-2 , 4 ).$

This is the fourth 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 $y=a$
  • Shifting of origin from (0,0) to $(h,k)$ then if the new axes form $X-Y$ coordinate system then
  • $x=X+a$
  • $y=Y+k$
  • The new coordinates of a point $P(x,y)$ become $X=x-a$,$Y=y-k$.
Step 1:
$VF=1+2=3$units=$a$
$VF$ is the axis whose equation is $y=4$(parallel to the $x$-axis).
Step 2:
Focus is on the -ve side of the vertex.
Therefore equation is of the form $y^2=-4ax$
Where $X=(x-1),Y=(y-4),a=3$
$(y-4)^2=-4\times 3(x-1)$
$\Rightarrow -12(x-1)$
$(y-4)^2=-12(x-1)$
This is the required equation.
answered Jun 14, 2013 by sreemathi.v
edited Jun 23, 2013 by sreemathi.v
 
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