Question

One end of latusrectum of the parabola is

Answer 1

$\huge{\underline{\boxed{\tt{\red{Answer}}}}}$

Given equation is y2−4x−2y−3=0

⇒(${y}^2$ −2y)=4x+3

⇒${(y-1)}^2$ −1=4x+3

⇒$({y-1)}^2$ =4(x+1)

Shift the origin to (−1,1)

⇒${y}^2$ =4X

Here focus is at (1,0).

Hence, focus of original parabola becomes

(1−1,0+1)=(0,1)

Therefore, equation of latusrectum is x=0.

Thus,

Point of intersection of parabola and latus rectum is,

${y}^2$ −2y−3=0

⇒ y = −1 or 3

So, the required points are (0,−1),(0,3).