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Focus of x²+y²+2xy-6x-2y+3=0 | JEE Rotated Parabola

JEE Maths question with a full step-by-step solution.

Question
The focus of the parabola x2+y2+2xy−6x−2y+3=0x^2 + y^2 + 2xy - 6x - 2y + 3 = 0 is:
A(1,−1)(1, -1)
B(−1,1)(-1, 1)
C(3,1)(3, 1)
D(1,1)(1, 1)correct
Solution
Step 1: Group the square (x+y)2=6x+2y−3(x + y)^2 = 6x + 2y - 3, then add 2λ(x+y)+λ22\lambda(x+y)+\lambda^2 and force the right side perpendicular to x+y+λ=0x + y + \lambda = 0:
2λ+62λ+2=−1⇒λ=−2⇒(x+y−2)2=2x−2y+1.\frac{2\lambda + 6}{2\lambda + 2} = -1 \Rightarrow \lambda = -2 \Rightarrow (x + y - 2)^2 = 2x - 2y + 1.
 Step 2: With Y=x+y−22Y = \dfrac{x + y - 2}{\sqrt{2}} and X=2x−2y+122X = \dfrac{2x - 2y + 1}{2\sqrt{2}} this is Y2=2 XY^2 = \sqrt{2}\,X, so 4A=24A = \sqrt{2}. The focus is Y=0, X=AY = 0,\ X = A:
x+y−2=0,2x−2y+1=1⇒x=y=1.x + y - 2 = 0,\quad 2x - 2y + 1 = 1 \Rightarrow x = y = 1.
 Correct answer: (4)
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