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Find the equation of parabola if : Vertex $(1 , 4); $ open left ward and passing through the point $(-2,10)$

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

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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:
Vertex $V(1,4)$ open left ward,passing through $P(-2,10)$
Since the parabola opens leftwards,its equation is of the form $Y^2=4aX$
(I.e)$(y-4)^2=-4a(x-1)$
Step 2:
It passes through $(-2,10)$
Therefore $(10-4)^2=-4a(-2-1)$
$\Rightarrow 6^2=12a$
$\Rightarrow 36=12a$
$a=3$
Therefore the equation is $(y-4)^2=-12(x-1)$
answered Jun 24, 2013 by sreemathi.v
 
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