Question

The coordinates of the end point of the latus rectum of the parabola ${(y-1)}^2=2(x+2)$, which does not lie on the line $2x+y+3=0$ are

• A. $(-2,1)$
• B. $(-\frac{3}{2},1)$
• C. $(-\frac{3}{2},2)$
• D. $(-\frac{3}{2},0)$

Answer 1

Answer: (C)

$(-\frac{3}{2},2)$

We know coordinate of the ends of the latus rectum of parabola $Y^2=4aX$ are given by,
$X=a,Y=2a$ and $X=a,Y=-2a$
Here $X=x+2,Y=y-1$ and $a=\frac{1}{2}$
Thus ends of the latus rectum of the given parabola are $(-\frac{3}{2},0)$ and $(-\frac{3}{2},2)$
Clearly point $(-\frac{3}{2},0)$ lies on the line $2x+y+3=0$