# 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

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.