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# Find the equation of parabola if : Vertex $(1 , 2 ) ;$ latus recturm :$y=5$.

This is the fifth 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 $V(1,2)$,latus rectum $y=5$
$F$ lies on the latus rectum, $VF$ perpendicular to the latus rectum.
Step 2:
$F$ is the foot of the $\perp$ from $V$ on the line $Y=5$.
Therefore $F(1,5)$
$VF=3$units=$a$
The parabola opens upwards.
Therefore the equation is $X^2=4aY$----(1)
Where $X=(x-1),Y=(y-2),a=3$
Substitute these in eq(1) we get
$(x-1)^2=4\times 3\times (y-2)$
$(x-1)^2=12(y-2)$
edited Jun 23, 2013