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how to prove that that length of focal chord of standard ellipse(a>b) which inclined angle titha to the major axis is 2ab^2/(a^2sin^2titha+b^2cos^2titha

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6 years ago

```							Equation of focal chord inclined at angle titha with x axis is y = tantitha(x-ae)Let it intersect x^2/a^2 + y^2/b^2 = 1 at P(x1.y1) & Q(x2,y2). Solving we getb^2*x^2 + a^2*tan^2titha*(x-ae)^2 = a^2*b^2This is a quadratic equation in x. It's roots are x1 & x2.abs(x1-x2) = sqrt( x1+x2)^2-4 x1 x2) =2ab^2 sectitha/b^2+a^2tan^2titha abs(y1-y2) =abs( (x1-ae)tantitha-(x2-ae) tantitha) = abs( (x1-x2)tantitha)Lenght of focal chord = PQ = sqrt( (x1-x2)^2+(y1-y2)^2)) = 2ab^2/b^2cos^2titha+a^2sin^2titha
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6 years ago
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