InterviewSolution
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InterviewSolution
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If the tangent and normal to a rectangular hyperbola cut off intercepts `a_(1)` and `a_(2)` on one axis and `b_(1)` and `b_(2)` on the other, thenA. `a_(1)a_(2)+b_(1)b_(2)=0`B. `a_(1)a_(2)=-b_(1)b_(2)`C. `a_(1)b_(2)=a_(2)b_(1)`D. `a_(1)a_(2)=b_(1)b_(2)` |
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Answer» Let the equation of the rectangular hyperbola be `x^(2)-y^(2)=a^(2)` and, let `P(a sec theta, a tan theta)` be a point on it. The equations of the tangent and normal at `P` are `x sec theta-y tantheta=a`……..`(i)` and, `x cos theta+y cot theta=2a`..........`(ii)` It is given that the tangent at `P` cuts off intercepts `a_(1)` and `b_(1)` on the coordinate axes. Therefore, `a_(1)=a cos theta` and `b_(1)=-a cot theta` The normal at `P` cuts off intercepts `a_(2)` and `b_(2)` on the coordinate axes. Therefore, `a_(2)=2a sec theta` and `b_(2)=2a tan theta` Now, `a_(1)a_(2)+b_(1)b_(2)=a cos thetaxx2a sec thet+(-a cot theta)(2a tantheta)` `impliesa_(1)a_(2)+b_(1)b_(2)=2a^(2)-2a^(2)=0` |
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