question_answer

The area of the rectangle formed by the perpendiculars from the centre of the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1(a>b>0)\] to the tangent and normal at its point whose eccentric angle is \[\frac{\pi }{4}\] is

A)  \[\frac{({{a}^{2}}-{{b}^{2}})ab}{{{a}^{2}}+{{b}^{2}}}\]                       

B)  \[\frac{({{a}^{2}}-{{b}^{2}})}{({{a}^{2}}+{{b}^{2}})ab}\]

C)  \[\frac{({{a}^{2}}-{{b}^{2}})}{ab({{a}^{2}}+{{b}^{2}})}\]                     

D)  \[\frac{({{a}^{2}}+{{b}^{2}})}{({{a}^{2}}-{{b}^{2}})ab}\]

Correct Answer: C

Solution :

\[P\left( \frac{a}{\sqrt{2}},\,\frac{b}{\sqrt{2}} \right){{P}_{1}}=\frac{\sqrt{2}ab}{{{a}^{2}}+{{b}^{2}}}\,;\,\,{{p}_{2}}=\frac{{{a}^{2}}-{{b}^{2}}}{\sqrt{2}\,({{a}^{2}}+{{b}^{2}})}\]\[\Rightarrow \,\,{{p}_{1}}{{p}_{2}}=\] result \[T:\,\,\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1\] \[{{p}_{1}}=\left| \frac{ab}{\sqrt{{{b}^{2}}{{\cos }^{2}}\,\theta +{{a}^{2}}{{\sin }^{2}}\theta }} \right|\,\]            ?(1) \[{{N}_{1}}=\frac{ax}{\cos \theta }\,-\frac{by}{\sin \theta }\,={{a}^{2}}-{{b}^{2}}\] \[{{p}_{2}}=\left| \frac{({{a}^{2}}-{{b}^{2}})\sin \theta \cos \theta }{\sqrt{{{a}^{2}}{{\sin }^{2}}\theta +{{b}^{2}}{{\cos }^{2}}\theta }}\, \right|\]    ?(2) \[{{p}_{1}}{{p}_{2}}=\frac{ab({{a}^{2}}-{{b}^{2}})}{2\left( \frac{{{a}^{2}}}{2}+\frac{{{b}^{2}}}{2} \right)}\]when \[\theta =\frac{\pi }{4};\,\,{{p}_{1}}{{p}_{2}}\,=\frac{ab({{a}^{2}}-{{b}^{2}})}{{{a}^{2}}+{{b}^{2}}}\]