Q35. If $x^{2} + 4 y^{2} = 4 x y$ , $t h e n, x : y i s$

(a) 2 : 2

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(b) 2 : 1

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(d) 1 : 4

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Solution

The correct option is B
(b) 2 : 1

$x^{2} + 4 y^{2} = 4 x y X^{2} - 4 x y + 4 y^{2} = 0 (x - 2 y)^{2} = 0 X = 2 y \frac{x}{y} = \frac{2}{1} x : y : : 2 : 1$

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