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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

x^2 + y^2 + 2xy - 6x - 2y + 3 = 0

is:

(1, -1)

(-1, 1)

(3, 1)

(1, 1)

correct

Solution

Step 1: Group the square

(x + y)^2 = 6x + 2y - 3

, then add

2\lambda(x+y)+\lambda^2

and force the right side perpendicular to

x + y + \lambda = 0

\frac{2\lambda + 6}{2\lambda + 2} = -1 \Rightarrow \lambda = -2 \Rightarrow (x + y - 2)^2 = 2x - 2y + 1.

Step 2: With

Y = \dfrac{x + y - 2}{\sqrt{2}}

and

X = \dfrac{2x - 2y + 1}{2\sqrt{2}}

this is

Y^2 = \sqrt{2}\,X

, so

4A = \sqrt{2}

. The focus is

Y = 0,\ X = A

x + y - 2 = 0,\quad 2x - 2y + 1 = 1 \Rightarrow x = y = 1.

Correct answer: (4)

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