Find the area of the rectangle formed by the perpendiculars from the center of the ellipse x 2 - a 2 - y 2 - b 2 -1 to the tangent and normal at a point whose eccentric angle is -4-

The correct option is A #160;( a ^ 2 b ^ 2 ) ab / a ^ 2 + b ^ 2 ...

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Logarithmic Function

Find the area...

Question

Find the area of the rectangle formed by the perpendiculars from the center of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ to the tangent and normal at a point whose eccentric angle is $\frac{π}{4} .$

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

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

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

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

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Solution

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2 a

2 b

\frac{π}{4}

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

x^{2} - y^{2} = 0

p (θ), D (θ + \frac{π}{2})

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

\frac{x^{2}}{a^{2} + b^{2}} + \frac{y^{2}}{b^{2}} = 1, \frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} + b^{2}} = 1

\frac{x^{2}}{a^{2} + b^{2}} + \frac{y^{2}}{b^{2}} = 1

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{a^{2} + b^{2}} = 1

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