InterviewSolution
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InterviewSolution
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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 |
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Answer» `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)` `=-x_1/y_1(1-e^2)` `m_T=-a/(Ke)(1-e^2)` `m_(PS)=(Ke)/(a(1-e^2))` `m_T=(-a(1-e^2))/(Ke)` `m_(PS)*m_(T)=-1` PS is perpendicular to tangent. Option B is correct. |
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