$$(\frac{1}{2}(a-b))^2+ab=p(a+b)^2$$ then the value of p is (assume that a ≠ - b)
$$(\frac{1}{2}(a-b))^2+ab=p(a+b)^2$$
The given expression can be written as $$((a-b))^2+4ab=4p(a+b)^2$$
$$a^2+b^2-2ab+4ab = 4p(a+b)^2$$
$$a^2 + b^2 + 2ab = 4p(a+b)^2$$
$$(a+b)^2 = 4p(a+b)^2$$
Hence, p = 1/4
Option D is the right answer.
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