KEAM 2015 Math Paper

y2−4x−2y−3=0
⇒‌‌(y2−2y)=4x+3
⇒‌‌(y−1)2−1=4x+3
⇒‌‌(y−1)2=4(x+1)
Shift the origin to (-1,1)
⇒‌‌Y2=4X
⇒ Focus is at (1,0)
Hence, focus of original parabola becomes (1−1,0+1)=(0,1)
∴ Equation of latusrectum is x=0.
∴ Point of intersection of parabola and latusrectum is
y2−2y−3=0⇒y=−1 or 3
So, required points are (0,−1),(0,3)