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Let P be any point on a directrix of an ellipse of eccentricity e, S be the corresponding focus and C be the centre of the ellipse. The line PC meets the ellipse at A. The angle between PS and tangent at A is `alpha`. Then `alpha` is equal to

Text Solution

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`m_(PS)=(K-0)/((a/e)-ae)`
`=K/(a((1-e^2)/e)`
`=(Ke)/(a(1-e^2))`
`(x x_1)/a^2+(yy_1)/b^2=1`
`m_T=-x_1/a^2*b^2/y_1`
`=x_1/y_1*(a^2(1-e^2))/a^2`
`Ax+By+C=0`
`m=-(A/B)`
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