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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)

satisfying

(2x + 1)^2 + y^2 + (y - 2x)^2 = \dfrac{1}{3}

(x, y) = \left(\dfrac{1}{3},\ -\dfrac{1}{3}\right)

(x, y) = \left(-\dfrac{1}{3},\ -\dfrac{1}{3}\right)

.correct

(x, y) = \left(-\dfrac{2}{3},\ -\dfrac{1}{3}\right)

(x, y) = \left(-\dfrac{1}{3},\ \dfrac{4}{3}\right)

Solution

Step 1: Multiply by

3

and expand.

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

2

12x^2 + 3y^2 - 6xy + 6x + 1 = 0.

Step 3: Regroup into squares.

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 = 0

and

3x + 1 = 0

. Hence

x = -\dfrac{1}{3}

and

y = x = -\dfrac{1}{3}

. Answer:

(x, y) = \left(-\dfrac{1}{3},\ -\dfrac{1}{3}\right)

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