Question : If $px + qy = 1$ and $q x+p y=\frac{2 p q}{p^2+q^2}$, then $\left(x^2+y^2\right)\left(p^2+q^2\right)$ is equal to:
Option 1: 2
Option 2: 3
Option 3: 0
Option 4: 1
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Correct Answer: 1
Solution :
Given: $px + qy = 1$---(1)
$q x+p y=\frac{2 p q}{p^2+q^2}$---(2)
Adding 1 and 2, we get,
$px+qy+qx+py=1+\frac{2 p q}{p^2+q^2}$
$⇒p(x+y)+q(x+y)=\frac{p^2+q^2+2 p q}{p^2+q^2}$
$⇒(x+y)(p+q)=\frac{(p+q)^2}{p^2+q^2}$
$⇒(x+y)=\frac{(p+q)}{p^2+q^2}$ ----------(3)
Subtracting 2 from 1, we get,
$px+qy-qx-py=1-\frac{2 p q}{p^2+q^2}$
$⇒p(x-y)-q(x-y)=\frac{p^2+q^2-2 p q}{p^2+q^2}$
$⇒(x-y)(p-q)=\frac{(p-q)^2}{p^2+q^2}$
$⇒(x-y)=\frac{(p-q)}{p^2+q^2}$ ---------(4)
Adding (3) and (4),
$⇒ 2x = \frac{2p}{p^2+q^2}$
$⇒ x = \frac{p}{p^2+q^2}$
$⇒ x^2 = \frac{p^2}{(p^2+q^2)^2}$------------(5)
Subtracting 4 from 3,
$⇒ 2y = \frac{2q}{p^2+q^2}$
$⇒ y = \frac{q}{p^2+q^2}$
$⇒ y^2 = \frac{q^2}{(p^2+q^2)^2}$------------(6)
Adding 5 and 6,
$⇒ x^2+y^2 = \frac{p^2+q^2}{(p^2+q^2)^2}$
$\left(x^2+y^2\right)\left(p^2+q^2\right) = \frac{p^2+q^2}{p^2+q^2}=1$
Hence, the correct answer is 1.
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