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The area of the rectangle formed by the perpendiculars from the centre of the ellipse to the tangent and normal at the point-whose eccentric angle is \pi / 4 \text {, is }

Option: 1

\mathrm{\left(\frac{a^2-b^2}{a^2+b^2}\right) a b}


Option: 2

\mathrm{\left(\frac{a^2+b^2}{a^2-b^2}\right) a b}


Option: 3

\mathrm{\frac{1}{a b}\left(\frac{a^2-b^2}{a^2+b^2}\right) a b}


Option: 4

\mathrm{\frac{1}{a b}\left(\frac{a^2+b^2}{a^2-b^2}\right) a b}


Answers (1)

best_answer

The given point is \mathrm{(a \cos \pi / 4, b \sin \pi / 4) \text { i.e. }} \mathrm{\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)}
So, the equation of the tangent at this point is \mathrm{\frac{x}{a}+\frac{y}{b}=\sqrt{2}}
∴ length of the perpendicular form (0, 0) on (i)  \mathrm{=\left|\frac{\frac{0}{a}+\frac{0}{b}-\sqrt{2}}{\sqrt{1 / a^2+1 / b^2}}\right|=\frac{\sqrt{2} a b}{\sqrt{a^2+b^2}}}
Equation of the normal at \mathrm{\left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)_{\text {is }} \frac{a^2 x}{a / \sqrt{2}}-\frac{b^2 y}{b / \sqrt{2}}=a^2-b^2 \Rightarrow \sqrt{2} a x-\sqrt{2} b y=a^2-b^2}
Therefore, length of the perpendicular form (0, 0) on (ii)
\mathrm{=\frac{a^2 b^2}{\sqrt{(\sqrt{2 a})^2+(-\sqrt{2 b})^2}}=\frac{a^2-b^2}{\sqrt{2\left(a^2+b^2\right)}}}
So, area of the rectangle
\mathrm{=p_1 p_2=\frac{\sqrt{2} a b}{\sqrt{a^2+b^2}} \times \frac{a^2-b^2}{\sqrt{2\left(a^2+b^2\right)}}=\left(\frac{a^2-b^2}{a^2+b^2}\right) a b}



 

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