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  • The area of the rectangle formed by the perpendicular drawn from the centre of the ellipse  <img alt="\mathrm{\frac{x^{2}}{a^{2}}+\frac{y^{a}}{b^{2}}=1}" src="https://entrancecorner.oncodecogs

The area of the rectangle formed by the perpendicular drawn from the centre of the ellipse  \mathrm{\frac{x^{2}}{a^{2}}+\frac{y^{a}}{b^{2}}=1} to the tangent and the normal at a point \mathrm{P} whose eccentric angle is \mathrm{\frac{\pi}{4}}

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)}
 


Option: 4

\text{ none of these}


Answers (1)

best_answer

Point \mathrm{P\: is \: \left(a \cos \frac{\pi}{4}, b \sin \frac{\pi}{4}\right)}

\mathrm{\Rightarrow \quad P\: is \: \left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right)}

Tangent at \mathrm{ P\: is\: \frac{x}{a}+\frac{y}{b}=\sqrt{2}}          ................(1)

The length of perpendicular on tangent

\mathrm{ =\left|\frac{0+0-\sqrt{2}}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}}}\right| }

\mathrm{ =\frac{\sqrt{2} a b}{\sqrt{a^2+b^2}} }

The equation of the normal at \mathrm{ \left(\frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}}\right) } is

\mathrm{ \frac{a^2 x}{a / \sqrt{2}}-\frac{b^2 y}{b / \sqrt{2}}=a^2-b^2 }

\mathrm{ Or \quad \sqrt{2} a x-\sqrt{2} b y=a^2-b^2 }          ..............(2)

\mathrm{ \therefore \quad } The length of perpendicular from \mathrm{ (0,0) } on normal at \mathrm{ P }

\mathrm{ =p_2=\frac{a^2-b^2}{\sqrt{2 a^2+2 b^2}} }

Area of the rectangle

\mathrm{ \therefore \quad=p_1 p_2 =\frac{\sqrt{2} a b}{\sqrt{a^2+b^2}} \frac{a^2-b^2}{\sqrt{2} \sqrt{a^2+b^2}} }

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

Hence option 2 is correct.





 

Posted by

seema garhwal

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