The area of the rectangle formed by the perpe | vedclass

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Question

$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}\

Vedclass · Accepted Answer

$\frac{{\left( {{a^2}\,\, - \,\,{b^2}} \right)\,\,ab}}{{{a^2}\,\, + \,\,{b^2}}}$

Vedclass · Answer

$P \left( {\frac{a}{{\sqrt 2 }}\,\,,\,\,\frac{b}{{\sqrt 2 }}} ight) p_1 = \frac{{\sqrt 2 \,\,ab}}{{{a^2}\,\, + \,\,{b^2}}}; p_2 = \frac{{{a^2}\,\, - \,\,{b^2}}}{{\sqrt 2 \,\,\left( {{a^2}\,\, + \,\,{b^2}} ight)}}$ $\Rightarrow p_1p_2 = $ result 
$ T : \frac{{x\,\cos 	heta }}{a}\,\, + \,\,\frac{{y\sin 	heta }}{b}\,\, = \,\,1$ 
$p_1 = \left| {\frac{{ab}}{{\sqrt {{b^2}{{\cos }^2}	heta \, + {a^2}{{\sin }^2}	heta } }}\,} ight|\,$ $ ....(1)$ 
$N_1 : \frac{{ax}}{{\cos 	heta }}\,\, - \,\,\frac{{by}}{{\sin 	heta }}\,\, = \,\,{a^2} - {b^2}$ 
$p_2 =\left| {\frac{{({a^2} - {b^2})\sin 	heta \cos 	heta }}{{\sqrt {{a^2}{{\sin }^2}	heta  + {b^2}{{\cos }^2}	heta } }}\,} ight|$ $ ....(2)$
$p_1p_2 = \frac{{ab({a^2} - {b^2})}}{{2\left( {\frac{{{a^2}}}{2}\, + \,\frac{{{b^2}}}{2}} ight)}}\,$ when $	heta = \pi /4;$ $p_1p_2 =\frac{{ab({a^2} - {b^2})}}{{{a^2} + {b^2}}}\,$

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 :

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