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# Find the equation of parabola if :Vertex $(3 , -1 );$ open rightward ; the distance between the latus recturm and directrix is $4$.

This is the eighth 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:
The parabola opens rightwards,the focus $F$ is to the right of the vertex and the latus rectum is to the left.
Since $V$ lies on the parabola,distance of $V$ from the latus rectum $VZ=VF=a$ units.
Step 2:
$ZF=2a=4$
The latus rectum is the chord through $F$ $\perp$ to $VF$
Therefore $a=2$ units.
The equation is of the form $Y^2=4aX$----(1)
Where $Y=y+1,X=(x-3),a=2$
Substitute these in eq(1) we get
$(y+1)^2=4\times 2\times (x-3)$
$\Rightarrow 8(x-3)$
$(y+1)^2=8(x-3)$
This is the required equation.
edited Jun 17, 2013