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 \[\frac{1}{{{p}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}\]You need to login to perform this action.
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