UPSC NDA Math Model Paper 3

Given equation: y2−8x+6y+1=0
⇒y2+6y+9−9−8x+1=0
⇒(y+3)2−8x−8=0
⇒(y+3)2=8x+8
⇒(y+3)2=8(x+1)
Let new coordinate axes be X and Y
Here X=x+1 and Y=y+3
⇒Y2=4aX
Now comparing with above equation,
∴4a=8⇒a=2
Focus: (a, 0)
X = a and Y = 0
⇒ x + 1 = 2 and y + 3 = 0
⇒ x = 1 and y = -3
∴ focus of parabola is (1, -3)