InterviewSolution
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InterviewSolution
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If the base of a triangle and the ratio of the lengths of the other two unequal sides are given, then the vertex lies onA. straight lineB. circleC. ellipseD. parabola |
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Answer» Correct Answer - B Let the base of the triangle be the line segment joining B(0, 0) and C(a, 0) and let the vertex be A(h, k), where a is fixed. Also, let `(AB)/(AC)=lambda, lambda !=1` `rArr AB^(2)=lambda^(2)AC^(2)` `rArrh^(2)+k^(2)=lambda^(2){(h-a)^(2)+k^(2)}` `rArr h^(2)(1-lambda^(2))+k^(2)(1-lambda^(2))+2a lambda^(2)h-a^(2)lambda^(2)=0` So, A(h, k) lies on `x^(2)+y^(2)+2a(lambda^(2))/(1-lambda^(2))x-(a^(2)lambda^(2))/(1-lambda^(2))=0`, which is a circle. |
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