The area of the rectangle formed by the perpe | vedclass

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Question

(A) Let the point on the ellipse be $P(a \cos 	heta, b \sin 	heta)$. For $	heta = \pi /4$,$P = (a/\sqrt{2}, b/\sqrt{2})$.The equation of the tangen

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the point on the ellipse be $P(a \cos 	heta, b \sin 	heta)$. For $	heta = \pi /4$,$P = (a/\sqrt{2}, b/\sqrt{2})$.The equation of the tangent at $P$ is $\frac{x \cos 	heta}{a} + \frac{y \sin 	heta}{b} = 1$.The perpendicular distance $p_1$ from the centre $(0,0)$ to the tangent is $p_1 = \frac{1}{\sqrt{\frac{\cos^2 	heta}{a^2} + \frac{\sin^2 	heta}{b^2}}} = \frac{ab}{\sqrt{b^2 \cos^2 	heta + a^2 \sin^2 	heta}}$.For $	heta = \pi /4$,$p_1 = \frac{ab}{\sqrt{b^2/2 + a^2/2}} = \frac{\sqrt{2} ab}{\sqrt{a^2 + b^2}}$.The equation of the normal at $P$ is $\frac{ax}{\cos 	heta} - \frac{by}{\sin 	heta} = a^2 - b^2$.The perpendicular distance $p_2$ from the centre $(0,0)$ to the normal is $p_2 = \frac{|a^2 - b^2|}{\sqrt{\frac{a^2}{\cos^2 	heta} + \frac{b^2}{\sin^2 	heta}}} = \frac{|a^2 - b^2| \sin 	heta \cos 	heta}{\sqrt{a^2 \sin^2 	heta + b^2 \cos^2 	heta}}$.For $	heta = \pi /4$,$p_2 = \frac{|a^2 - b^2| (1/2)}{\sqrt{a^2/2 + b^2/2}} = \frac{|a^2 - b^2|}{\sqrt{2} \sqrt{a^2 + b^2}}$.The area of the rectangle is $p_1 	imes p_2 = \left( \frac{\sqrt{2} ab}{\sqrt{a^2 + b^2}} ight) \left( \frac{|a^2 - b^2|}{\sqrt{2} \sqrt{a^2 + b^2}} ight) = \frac{ab(a^2 - b^2)}{a^2 + b^2}$.

The area of the rectangle formed by the perpendiculars from the centre of the standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ to the tangent and normal at its point whose eccentric angle is $\pi /4$ is:

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