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The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse to the tangent and normal at its point whose eccentric angle is $\pi /4$ is :
$\frac{{\left( {{a^2}\,\, - \,\,{b^2}} \right)\,\,ab}}{{{a^2}\,\, + \,\,{b^2}}}$
$\frac{{\left( {{a^2} - {b^2}} \right)}}{{\left( {{a^2} + {b^2}} \right)ab}}$
$\frac{{\left( {{a^2}\,\, - \,\,{b^2}} \right)}}{{ab\,\,\left( {{a^2}\,\, + \,\,{b^2}} \right)}}$
$\frac{{{a^2}\,\, + \,\,{b^2}}}{{\left( {{a^2}\,\, - \,\,{b^2}} \right)\,\,ab}}$
Solution
$P \left( {\frac{a}{{\sqrt 2 }}\,\,,\,\,\frac{b}{{\sqrt 2 }}} \right) p_1 = \frac{{\sqrt 2 \,\,ab}}{{{a^2}\,\, + \,\,{b^2}}}; p_2 = \frac{{{a^2}\,\, – \,\,{b^2}}}{{\sqrt 2 \,\,\left( {{a^2}\,\, + \,\,{b^2}} \right)}}$ $\Rightarrow p_1p_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)$
$p_1p_2 = \frac{{ab({a^2} – {b^2})}}{{2\left( {\frac{{{a^2}}}{2}\, + \,\frac{{{b^2}}}{2}} \right)}}\,$ when $\theta = \pi /4;$ $p_1p_2 =\frac{{ab({a^2} – {b^2})}}{{{a^2} + {b^2}}}\,$
