Question

If $2p$ is the length of the perpendicular from the origin to the lines $\frac{x}{a}+\frac{y}{b}=1$, then $a^2,8p^2,b^2$ are in

• A. AP
• B. GP
• C. HP
• D. None of these

Answer 1

The correct option is C

HP

Explanation of the correct answer:

Step 1: Prerequisite

It is known that the perpendicular distance of any line $ax+by+c=0$ from any point $(x_0,y_0)$ is given by

$d=\frac{\left|a{x}_0+b{y}_0+c_0\right|}{\sqrt{a^2+b^2}}$ (i)

Step 2: Evaluating the given data

From the question, the point can be assumed to be $(0,0)$ as $2p$ is the length from the origin to the lines.

The equation of the lines is given as $\frac{x}{a}+\frac{y}{b}=1$.

Step 3: Substituting the values and simplifying:

On substituting the values in equation (i), we get

$d=\frac{\left|a{x}_0+b{y}_0+c_0\right|}{\sqrt{a^2+b^2}}$

$2p=\frac{\left|a\left(0\right)+b\left(0\right)-1\right|}{\sqrt{{\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{b^2}}}}$

$2p=\frac{1}{\sqrt{{\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{b^2}}}}$

On squaring both sides, we obtain

$4p^2=\frac{1}{{\displaystyle \frac{1}{a^2}+\frac{1}{b^2}}}$

Taking reciprocals on both sides,

$\frac{1}{4p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

$\frac{2}{8p^2}=\frac{1}{a^2}+\frac{1}{b^2}$

Therefore, $a^2,8p^2,b^2$ are in HP.

Hence, Option (3) is correct.