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If $p$ is the length of the perpendicular from the origin on the line $\frac{x}{a} + \frac{y}{b} = 1$ and $a^{2}, p^{2}, b^{2}$ are in A.P., then $a^{4} - 2 p^{2} a^{2} + 2 p^{4} =$

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Correct option is B

We have p=|ab|a2+b2 ⇒ p2=a2b2a2+b2Also, since a2 , p2 , b2 are in A.P., a2+b2=2p2 and a2b2=p2a2+b2=2p4⇒ 2p4=a22p2−a2 ⇒ a4−2p2a2+2p4=0

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If p is the length of the perpendicular from the origin on the line xa+yb=1 and a2 , p2 , b2 are in A.P., then a4–2p2a2+2p4=

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Ready to Test Your Skills?

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If $p$ is the length of the perpendicular from the origin on the line $\frac{x}{a} + \frac{y}{b} = 1$ and $a^{2}, p^{2}, b^{2}$ are in A.P., then $a^{4} - 2 p^{2} a^{2} + 2 p^{4} =$