Question

What will be the equation/s of parabola with latus rectum joining the points $(3,6)$ and $(3,-2)$?

• A. $(y-2{)}^2=-8(x-1)$
• B. $(y-2{)}^2=8(x-1)$
• C. $(y+6{)}^2=-8(x-3)$
• D. $(y-2{)}^2=8(x-3)$

Answer 1

Answer: (A), (B)

$(y-2{)}^2=8(x-1)$

First, checking the slope of $(3,6)$ and $(3,-2)$ we get,
$\frac{6+2}{3-3}=\infty$
$\because$ latus rectum is perpendicular to axis.
Hence, axis parallel to x-axis the equation of two possible parabolas will be of form
$(y-k{)}^2=\pm 4a(x-h)$ $...\left(1\right)$
Since, latus rectum$=\sqrt{(3-3{)}^2+(6+2{)}^2}$
$\Rightarrow 4a=8\Rightarrow a=2.$
Also, the mid point of end points will be focus of these parabola i.e $(\frac{3+3}{2},\frac{6-2}{2})$
$\Rightarrow (3,2).$
On comparing with coordinates of focus of translated parabola $(a+h,k)$ we get $a+h=3\Rightarrow 2+h=3\Rightarrow h=1$ and $k=2.$
So, the vertex $(h,k)$ is $(1,2).$
Putting value of $(h,k)$ in equation $\left(1\right)$ we get,
$(y-2{)}^2=\pm 8(x-1).$