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(C) The equation of the line is $\frac{x}{a} + \frac{y}{b} - 1 = 0$.The length of the perpendicular from the origin $(0,0)$ to the line is given by $d

Vedclass · Accepted Answer

$a^2, 8p^2, b^2$ are in $H.P.$

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(C) The equation of the line is $\frac{x}{a} + \frac{y}{b} - 1 = 0$.The length of the perpendicular from the origin $(0,0)$ to the line is given by $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.Here,$2p = \frac{|-1|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}}$.Squaring both sides,$4p^2 = \frac{1}{\frac{1}{a^2} + \frac{1}{b^2}}$,which implies $\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4p^2}$.Multiplying by $2$,we get $\frac{2}{a^2} + \frac{2}{b^2} = \frac{2}{4p^2} = \frac{1}{2p^2}$.Wait,the standard relation is $\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$ where $p$ is the perpendicular. Given $2p$ is the length,let $L = 2p$. Then $\frac{1}{L^2} = \frac{1}{a^2} + \frac{1}{b^2}$.So,$\frac{1}{4p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.This means $\frac{1}{a^2}, \frac{1}{4p^2}, \frac{1}{b^2}$ are in $A.P.$ if we consider the harmonic mean property,but specifically,$\frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{4p^2}$.Actually,$\frac{1}{a^2}, \frac{1}{4p^2}, \frac{1}{b^2}$ are in $A.P.$ implies $\frac{2}{4p^2} = \frac{1}{a^2} + \frac{1}{b^2}$,which is $\frac{1}{2p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.Given $\frac{1}{4p^2} = \frac{1}{a^2} + \frac{1}{b^2}$,we have $\frac{1}{a^2}, \frac{1}{8p^2}, \frac{1}{b^2}$ in $A.P.$ implies $\frac{2}{8p^2} = \frac{1}{a^2} + \frac{1}{b^2}$,which is $\frac{1}{4p^2} = \frac{1}{a^2} + \frac{1}{b^2}$.Thus,$\frac{1}{a^2}, \frac{1}{8p^2}, \frac{1}{b^2}$ are in $A.P.$,which implies $a^2, 8p^2, b^2$ are in $H.P.$

If $2p$ is the length of the perpendicular from the origin to the line $\frac{x}{a} + \frac{y}{b} = 1$,then:

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