Question

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 $ab$ is equal to

• A. $p^2$
• B. $\sqrt{2}{p}^2$
• C. $-\sqrt{2}{p}^2$
• D. $2p^2$

Answer 1

Answer: (B), (C)

The correct options are
B $-\sqrt{2}{p}^2$
C $\sqrt{2}{p}^2$

Given $a^2,p^2,b^2$ are in A.P

Hence $p^2=\frac{a^2+b^2}{2}$----(1)

p is the perpendicular distance of given line $bx+ay=ab$

$p=\left|\frac{-ab}{\sqrt{a^2+b^2}}\right|$

$p=\left|\frac{-ab}{\sqrt{2p^2}}\right|$ from eq (1)

$p=\left|\frac{ab}{p\sqrt{2}}\right|$

$\pm p^2\sqrt{2}=ab$

$ab=+p^2\sqrt{2},-p^2\sqrt{2}$