Question

The one end of the latus rectum of the parabola $y^2-4x-2y-3=0$ is at

• A. (0,-1)
• B. (0,1)
• C. (0,-3)
• D. (3, 0)
• E. (0,2)

Answer 1

Answer: (A)

$y^2-4x-2y-3=0$
$\Rightarrow \left(y^2-2y\right)=4x+3$
$\Rightarrow (y-1{)}^2-1=4x+3$
$\Rightarrow (y-1{)}^2=4(x+1)$
Shift the origin to $(-1,1)$. $\Rightarrow Y^2=4X$
$\Rightarrow$ Focus is at $(1,0)$.
Hence, focus of original parabola becomes $(1-1,0+1)=(0,1)$
$\therefore$ Equation of latusrectum is $x=0$
$\therefore$ Point of intersection of parabola and latusrectum is
$y^2-2y-3=0\Rightarrow y=-1$ or }$3$
So, required points are $(0,-1),(0,3)$.