question_answer

If p is the length of the perpendicular from the origin on the line whose intercepts on the axes are a and b then :

A)  \[{{p}^{2}}={{a}^{2}}+{{b}^{2}}\]                             

B)  \[{{p}^{2}}={{a}^{2}}-{{b}^{2}}\]

C)  \[\frac{1}{{{p}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}\]    

D)  \[\frac{1}{{{p}^{2}}}=\frac{1}{{{a}^{2}}}-\frac{1}{{{b}^{2}}}\]

Correct Answer: C

Solution :

Here the equation of AB is \[\frac{x}{a}+\frac{y}{b}=1\] From the figure \[OP\bot AB\]                 \[\therefore \]  \[OP=\left| \frac{0.\left( \frac{1}{a} \right)+0\left( \frac{1}{b} \right)-1}{\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}} \right|\]                                 \[p=\frac{-1}{\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}}\]                 \[{{p}^{2}}=\frac{1}{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}\] [Squaring both sides] or            \[\frac{1}{{{p}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}\]