A) \[\frac{\left( {{a}^{2}}-{{b}^{2}} \right)ab}{{{a}^{2}}+{{b}^{2}}}\]
B) \[\frac{\left( {{a}^{2}}+{{b}^{2}} \right)ab}{{{a}^{2}}-{{b}^{2}}}\]
C) \[ab\]
D) \[\frac{{{a}^{2}}{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\]
Correct Answer: A
Solution :
| Equation of the tangent at \[\frac{\pi }{4}\]is \[\frac{x\left( \frac{1}{\sqrt{2}} \right)}{a}+\frac{y\left( \frac{1}{\sqrt{2}} \right)}{b}=1\] |
| i.e., \[\frac{x}{a}+\frac{y}{b}-\sqrt{2}=0..(i)\] |
| Equation of the normal at \[\frac{\pi }{4}\]is \[\frac{x}{a}-\frac{y}{b}=\frac{a}{b\sqrt{2}}-\frac{b}{a\sqrt{2}}...(ii)\] |
| \[{{P}_{1}}\]=length of the perpendicular from the Centre to the tangent \[=\left| \frac{-\sqrt{2}}{\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}} \right|\] |
| \[=\frac{\sqrt{2}ab}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\] |
| \[{{P}_{2}}\]=length of the perpendicular from the Centre to the normal |
| \[=\left| \frac{\frac{a}{b\sqrt{2}}-\frac{b}{a\sqrt{2}}}{\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}}} \right|\]\[=\frac{{{a}^{2}}-{{b}^{2}}}{\sqrt{2}\sqrt{{{a}^{2}}+{{b}^{2}}}}.\] |
| Area of the rectangle =\[{{P}_{1}}{{P}_{2}}\]\[=\frac{ab\left( {{a}^{2}}-{{b}^{2}} \right)}{{{a}^{2}}+{{b}^{2}}}\] |
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