A) straight line
B) circle
C) ellipse
D) parabola
Correct Answer: B
Solution :
Let the base of the triangle be the line joining B (0, 0)and C (a, 0) and let the third vertex be A (h, k), where a is fixed. Again, let \[\frac{AB}{AC}=\lambda ,\lambda \ne 1\] \[\Rightarrow \]\[A{{B}^{2}}={{\lambda }^{2}}A{{C}^{2}}\] \[\Rightarrow \]\[{{h}^{2}}+{{k}^{2}}={{\lambda }^{2}}\{{{(h-a)}^{2}}+{{k}^{2}}\}\] \[\Rightarrow \]\[{{h}^{2}}(1-{{\lambda }^{2}})+{{k}^{2}}(1-{{\lambda }^{2}})+2a{{\lambda }^{2}}h-{{a}^{2}}{{\lambda }^{2}}=0\] So, A (h, k) lies on \[{{x}^{2}}+{{y}^{2}}+2a\frac{{{\lambda }^{2}}}{1-{{\lambda }^{2}}}x-\frac{{{a}^{2}}{{\lambda }^{2}}}{1-{{\lambda }^{2}}}=0,\]which represents an equation of circle.
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