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Real Pairs with (2x+1)²+y²+(y-2x)²=1/3 | IOQM

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

Question
The all pairs of real numbers (x,y)(x, y) satisfying (2x+1)2+y2+(y−2x)2=13(2x + 1)^2 + y^2 + (y - 2x)^2 = \dfrac{1}{3}.
A(x,y)=(13, −13)(x, y) = \left(\dfrac{1}{3},\ -\dfrac{1}{3}\right).
B(x,y)=(−13, −13)(x, y) = \left(-\dfrac{1}{3},\ -\dfrac{1}{3}\right).correct
C(x,y)=(−23, −13)(x, y) = \left(-\dfrac{2}{3},\ -\dfrac{1}{3}\right).
D(x,y)=(−13, 43)(x, y) = \left(-\dfrac{1}{3},\ \dfrac{4}{3}\right).
Solution
Step 1: Multiply by 33 and expand.
3[(2x+1)2+y2+(y−2x)2]=1  ⟹  24x2+6y2−12xy+12x+2=0.3\big[(2x + 1)^2 + y^2 + (y - 2x)^2\big] = 1 \implies 24x^2 + 6y^2 - 12xy + 12x + 2 = 0.
 Step 2: Divide by 22:
12x2+3y2−6xy+6x+1=0.12x^2 + 3y^2 - 6xy + 6x + 1 = 0.
 Step 3: Regroup into squares.
3(x2−2xy+y2)+(9x2+6x+1)=0  ⟹  3(x−y)2+(3x+1)2=0.3(x^2 - 2xy + y^2) + (9x^2 + 6x + 1) = 0 \implies 3(x - y)^2 + (3x + 1)^2 = 0.
 Step 4: A sum of squares of real numbers is zero only if each is zero: x−y=0x - y = 0 and 3x+1=03x + 1 = 0. Hence x=−13x = -\dfrac{1}{3} and y=x=−13y = x = -\dfrac{1}{3}. Answer: (x,y)=(−13, −13)(x, y) = \left(-\dfrac{1}{3},\ -\dfrac{1}{3}\right).
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