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Parabola: Parabola Given 2xy

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

Question
A parabola is given by x2+y2+2xy−4x+4y−4=0x^{2} + y^{2} + 2xy - 4x + 4y - 4 = 0. Then
AIts focus is (0,0)(0,0)correct
BIts focus is (−12, 12)\left(-\dfrac12,\,\dfrac12\right)
CIts directrix is x−y+2=0x - y + 2 = 0correct
DIts directrix is x−y−2=0x - y - 2 = 0
Solution
Question attachment Step 1: Grouping the terms,
x2+2xy+y2=4x−4y+4 ⟹ (x+y)2=4(x−y+1)x^{2}+2xy+y^{2} = 4x - 4y + 4 \ \Longrightarrow\ (x+y)^{2} = 4(x-y+1)
 The lines x+y=0x+y=0 and x−y+1=0x-y+1=0 are perpendicular, so this is a parabola with a tilted axis. Step 2: Divide each linear form by 12+12=2\sqrt{1^{2}+1^{2}} = \sqrt2, so that it measures a distance. With (x+y)2=2(x+y2)2(x+y)^{2} = 2\left(\dfrac{x+y}{\sqrt2}\right)^{2} and x−y+1=2(x−y+12)x-y+1 = \sqrt2\left(\dfrac{x-y+1}{\sqrt2}\right),
2(x+y2)2=42(x−y+12) ⟹ (x+y2)2=22(x−y+12)2\left(\frac{x+y}{\sqrt2}\right)^{2} = 4\sqrt2\left(\frac{x-y+1}{\sqrt2}\right) \ \Longrightarrow\ \left(\frac{x+y}{\sqrt2}\right)^{2} = 2\sqrt2\left(\frac{x-y+1}{\sqrt2}\right)
 Step 3: This is Y2=4AXY^{2} = 4AX in the rotated frame
Y=x+y2,X=x−y+12,4A=22 ⟹ A=12Y = \frac{x+y}{\sqrt2}, \qquad X = \frac{x-y+1}{\sqrt2}, \qquad 4A = 2\sqrt2 \ \Longrightarrow\ A = \frac{1}{\sqrt2}
 Step 4: The vertex is X=Y=0X=Y=0, i.e. the intersection of x+y=0x+y=0 and x−y+1=0x-y+1=0:
x+y=0, x−y=−1 ⟹ x=−12, y=12x+y=0,\ x-y=-1 \ \Longrightarrow\ x = -\frac12,\ y = \frac12
 Step 5: The focus is at a distance AA from the vertex along the axis, in the direction of increasing XX, which is the unit normal 12(1,−1)\dfrac{1}{\sqrt2}(1,-1) of the line x−y+1=0x-y+1=0:
S=(−12,12)+12⋅12(1,−1)=(−12+12, 12−12)=(0,0)S = \left(-\frac12,\frac12\right) + \frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}(1,-1) = \left(-\frac12 + \frac12,\ \frac12 - \frac12\right) = (0,0)
 So (A) is true, and (2) names the vertex, not the focus, so it is false. Step 6: The directrix is X=−AX = -A, i.e.
x−y+12=−12 ⟹ x−y+1=−1 ⟹ x−y+2=0\frac{x-y+1}{\sqrt2} = -\frac{1}{\sqrt2} \ \Longrightarrow\ x-y+1 = -1 \ \Longrightarrow\ x - y + 2 = 0
 So (3) is true and (4) is false. Step 7: Check with the focus-directrix definition at the vertex: the distance from (−12,12)\left(-\tfrac12,\tfrac12\right) to (0,0)(0,0) is 12\dfrac{1}{\sqrt2}, and its distance from x−y+2=0x-y+2=0 is ∣−1/2−1/2+2∣2=12\dfrac{|-1/2-1/2+2|}{\sqrt2} = \dfrac{1}{\sqrt2}. Equal, as required. Answer: (1) and (3).
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