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(B) Let the coordinates of the moving point be $(x, y)$.According to the definition of an ellipse,the sum of the distances of a point from two fixed p

Vedclass · Accepted Answer

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

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(B) Let the coordinates of the moving point be $(x, y)$.According to the definition of an ellipse,the sum of the distances of a point from two fixed points (foci) is constant $(2a)$.The given condition is $\sqrt{(x - ae)^2 + y^2} + \sqrt{(x + ae)^2 + y^2} = 2a$.Squaring and simplifying this equation:$\sqrt{(x - ae)^2 + y^2} = 2a - \sqrt{(x + ae)^2 + y^2}$.Squaring both sides:$(x - ae)^2 + y^2 = 4a^2 + (x + ae)^2 + y^2 - 4a\sqrt{(x + ae)^2 + y^2}$.$x^2 - 2aex + a^2e^2 + y^2 = 4a^2 + x^2 + 2aex + a^2e^2 + y^2 - 4a\sqrt{(x + ae)^2 + y^2}$.$-4aex - 4a^2 = -4a\sqrt{(x + ae)^2 + y^2}$.$ex + a = \sqrt{(x + ae)^2 + y^2}$.Squaring again:$e^2x^2 + 2aex + a^2 = x^2 + 2aex + a^2e^2 + y^2$.$x^2(1 - e^2) + y^2 = a^2(1 - e^2)$.Dividing by $a^2(1 - e^2)$:$\frac{x^2(1 - e^2)}{a^2(1 - e^2)} + \frac{y^2}{a^2(1 - e^2)} = 1$.Since $b^2 = a^2(1 - e^2)$,we get $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

$A$ point moves such that the sum of its distances from $(ae, 0)$ and $(-ae, 0)$ is $2a$. Then the equation to its locus,where $b^2 = a^2(1 - e^2)$,is

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