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Maths-

General

Easy

Question

P and Q are two variable points on the axes of x and y respectively such that $open vertical bar O P close vertical bar plus open vertical bar O Q close vertical bar equals a$ , then the locus of the foot of the perpendicular from the origin on PQ is

$(x - y) (x^{2} + y^{2}) = a x y$
$(x + y) (x^{2} + y^{2}) = a x y$
$(x + y) (x^{2} + y^{2}) = a (x - y)$
$(x + y) (x^{2} - y^{2}) = a x y$

The correct answer is: $open parentheses x plus y close parentheses open parentheses x to the power of 2 end exponent plus y to the power of 2 end exponent close parentheses equals a x y$

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