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If the chord joining the points `(a sectheta_1, b tantheta_1)` and `(a sectheta_2, b tantheta_2)` on the hyperbola `x^2/a^2-y^2/b^2=1` is a focal chord, then prove that `tan(theta_1/2)tan(theta_2/2)+(ke-1)/(ke+1)=0`, where `k=+-1`A. `(1-e)/(1+e)`B. `(e-1)/(e+1)`C. `(e+1)/(e-1)`D. `(1+e)/(1-e)` |
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Answer» The equation of the chord joining `(a sectheta_(1), b tantheta_(1))` and `(a sec theta_(2), b tan theta_(2))` is `(x)/(a)cos((theta_(1)-theta_(2))/(2))-(y)/(b)sin((theta_(1)+theta_(2))/(2))=cos((theta_(1)+theta_(2))/(2))` If it passes through the focus `(ae,0)` then `e cos ((theta_(1)-theta_(2))/(2))=cos((theta_(1)+theta_(2))/(2))` `implies(cos((theta_(1)-theta_(2))/(2)))/(cos((theta_(1)+theta_(2))/(2)))=(1)/(e)` `implies(cos((theta_(1)-theta_(2))/(2))+cos((theta_(1)+theta_(2))/(2)))/(cos((theta_(1)+theta_(2))/(2))-cos((theta_(1)+theta_(2))/(2)))=(1+e)/(1-e)` `impliescot.(theta_(1))/(2)cot.(theta_(2))/(2)=(1+e)/(1-e)impliestan.(theta_(1))/(2)tan.(theta_(2))/(2)=(1-e)/(1+e)` |
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