question_answer

The line L has intercepts a and b on the coordinate axes. Keeping the origin fixed, the coordinate axes are rotated through a fixed angle. If the line L has intercepts p and q on the rotated axes, then\[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}\]is equal to

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,      \[{{p}_{1}}={{p}_{2}}\]                 \[=\frac{\left| \frac{0}{a}+\frac{0}{b}-1 \right|}{\sqrt{\frac{a}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}}=\frac{\left| \frac{0}{p}+\frac{0}{q}-1 \right|}{\sqrt{\frac{1}{{{p}^{2}}}+\frac{1}{{{q}^{2}}}}}\] \[\Rightarrow \]               \[\sqrt{\frac{1}{{{p}^{2}}}+\frac{1}{{{q}^{2}}}}=\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}\] \[\Rightarrow \]               \[\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{p}^{2}}}+\frac{1}{{{q}^{2}}}\]