InterviewSolution
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InterviewSolution
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Normal are drawn to the hyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1`at point `theta_1a n dtheta_2`meeting the conjugate axis at `G_1a n dG_2,`respectively. If `theta_1+theta_2=pi/2,`prove that `C G_1dotC G_2=(a^2e^4)/(e^2-1)`, where `C`is the center of the hyperbola and `e`is the eccentricity. |
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Answer» The normal at point `P(a sec theta_(1), b tan theta_(2))` is `ax cos theta_(1)+by cot theta_(1)=(a^(2)+b^(2))` It meets the conjugate axis at `G_(1)(0,(a^(2)+b^(2))/(b)tan theta_(1))`. The normal at point `Q(a sec theta_(2), b tan theta_(2))` is `axcos theta_(2)+" by " cot theta_(2)=(a^(2)+b^(2))` It meets the conjugate axis at `G_(1)(0,(a^(2)+b^(2))/(b)tan theta_(2)).` Therefore, `CG_(1)*CG_(2)=((a^(2+b^(2)))^(2))/(b^(2))tantheta_(1)tantheta_(2)` `=((a^(2)+b^(2)))/(b^(2))" "(becausetheta_(1)+theta_(2)=(pi)/(2))` `=(a^(4)(1+(b^(2))/(a^(2))))/(b^(2))` `=(a^(2)e^(4))/(e^(2)-1)` |
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