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Foci of the Rotated Ellipse 3x^2-4xy+3y^2 = 5 | JEE

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

Question
The foci of the ellipse 3x2−4xy+3y2=53x^2-4xy+3y^2=5 are
A(2,2),(−2,−2)(\sqrt2,\sqrt2),(-\sqrt2,-\sqrt2)correct
B(2,−2),(−2,2)(\sqrt2,-\sqrt2),(-\sqrt2,\sqrt2)
C(±2,0)(\pm2,0)
D(22,22),(−22,−22)(2\sqrt2,2\sqrt2),(-2\sqrt2,-2\sqrt2)
Solution
Step 1: Remove the xyxy-term by rotating the axes through 45∘45^\circ using x=X−Y2, y=X+Y2x=\dfrac{X-Y}{\sqrt2},\ y=\dfrac{X+Y}{\sqrt2}. Step 2: Substitute.
3(x2+y2)−4xy=3(X2+Y2)−2(X2−Y2)=X2+5Y2.3(x^2+y^2)-4xy=3(X^2+Y^2)-2(X^2-Y^2)=X^2+5Y^2.
 So the equation becomes X2+5Y2=5X^2+5Y^2=5, i.e. X25+Y21=1\dfrac{X^2}{5}+\dfrac{Y^2}{1}=1. Step 3: Here a2=5, b2=1a^2=5,\ b^2=1, so c2=a2−b2=4, c=2c^2=a^2-b^2=4,\ c=2. The foci in the rotated frame are (±2,0)(\pm2,0). Step 4: Transform back. For (2,0)(2,0): x=2−02=2, y=2+02=2x=\dfrac{2-0}{\sqrt2}=\sqrt2,\ y=\dfrac{2+0}{\sqrt2}=\sqrt2. For (−2,0)(-2,0): (−2,−2)(-\sqrt2,-\sqrt2). ∴\therefore foci are (2,2)(\sqrt2,\sqrt2) and (−2,−2)(-\sqrt2,-\sqrt2). Correct answer: (1)
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