InterviewSolution
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InterviewSolution
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Normal is drawn at one of the extremities of the latus rectum of thehyperbola `(x^2)/(a^2)-(y^2)/(b^2)=1`which meets the axes at points `Aa n dB`. Then find the area of triangle `O A B(O`being the origin). |
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Answer» Correct Answer - `"Area"=(1)/(2)a^(2)e^(5)` Normal at point `P(x_(1),y_(1))` is `(a^(2)x)/(x_(1))+(b^(2)y)/(y_(1))=a^(2)+b^(2).` It meets the axes at `A(((a^(2)_b^(2))x_(1))/(a^(2)),0)and B(0,((a^(2)+b^(2))y_(1))/(b^(2)))` `"Area of "DeltaOAB=(1)/(2)[((a^(2)+b^(2))x_(1))/(a^(2))][((a^(2)+b^(2))y_(1))/(b^(2))]` `=(1)/(2)[((a^(2)+b^(2))x_(1)y_(1))/(a^(2)b^(2))]` Now, normal is drawn at the extremity of latus rectum. Hence, `(x_(1),y_(1))-=(ae,(b^(2))/(a))` `therefore" Area"=(1)/(2)[((a^(2)+b^(2))^(2)b^(2)e)/(a^(2)b^(2))]` `=(1)/(2)[(a^(4)(1+(b^(2))/(a^(2)))^(2)e)/(a^(2))]` `=(1)/(2)a^(2)e^(5)` |
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