A(a,0) and B(-a,0) are two fixed points of tr | vedclass

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(A) Let the coordinates of $A$ and $B$ be $A(a, 0)$ and $B(-a, 0)$ and the variable point be $C(h, k)$.From the figure,$\cot A = \frac{a - h}{k}$ and

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$y\lambda = 2a$

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(A) Let the coordinates of $A$ and $B$ be $A(a, 0)$ and $B(-a, 0)$ and the variable point be $C(h, k)$.From the figure,$\cot A = \frac{a - h}{k}$ and $\cot B = \frac{a + h}{k}$.According to the condition,$\cot A + \cot B = \lambda$.Substituting the values,we get $\frac{a - h}{k} + \frac{a + h}{k} = \lambda$.This simplifies to $\frac{2a}{k} = \lambda$,which implies $k\lambda = 2a$.Replacing $(h, k)$ with $(x, y)$,the locus is $y\lambda = 2a$.

$A(a,0)$ and $B(-a,0)$ are two fixed points of triangle $ABC$. The vertex $C$ moves in such a way that $\cot A + \cot B = \lambda$,where $\lambda$ is a constant. Then the locus of the point $C$ is

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