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Q)

Find the equation of parabola if : Vertex $(3 , -2 ); $ open downward and the length of the latus rectum is $8$.

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

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A)
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 $(3,-2)$
Open downward and length of latus rectum is 8.
Latus rectum=$4a=8$
$\Rightarrow a=2$
Step 2:
The parabola opens downwards,the equation of the form $X^2=-4aY$-----(1)
Where $X=(x-3),Y=(y+2),a=2$
Substitute these in eq(1) we get
$(x-3)^2=-4\times 2\times (y+2)$
$\Rightarrow -8(y+2)$
$(x-3)^2=-8(y+2)$
This is the required equation.
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