A) \[{{p}^{2}}+{{q}^{2}}\]
B) \[{{p}^{2}}-{{q}^{2}}\]
C) \[\frac{1}{{{p}^{2}}}+\frac{1}{{{q}^{2}}}\]
D) \[\frac{1}{{{p}^{2}}}-\frac{1}{{{q}^{2}}}\]
E) \[\frac{1}{{{q}^{2}}}-\frac{1}{{{p}^{2}}}\]
Correct Answer: C
Solution :
Since the origin remains same, so the length of the perpendicular from the origin on the line in this positions \[\frac{x}{a}+\frac{y}{b}=1\] ...(i) and \[\frac{x}{p}+\frac{y}{q}=1\] ...(ii) are equal. Therefore, \[{{p}_{1}}={{p}_{2}}\] \[=\frac{\left| \frac{0}{a}+\frac{0}{b}-1 \right|}{\sqrt{\frac{a}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}}=\frac{\left| \frac{0}{p}+\frac{0}{q}-1 \right|}{\sqrt{\frac{1}{{{p}^{2}}}+\frac{1}{{{q}^{2}}}}}\] \[\Rightarrow \] \[\sqrt{\frac{1}{{{p}^{2}}}+\frac{1}{{{q}^{2}}}}=\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}\] \[\Rightarrow \] \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{p}^{2}}}+\frac{1}{{{q}^{2}}}\]You need to login to perform this action.
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