A point moves such that the sum of its distan | vedclass

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(A) Let $P(x, y)$ be the moving point and $F_1(ae, 0)$,$F_2(-ae, 0)$ be the fixed points.According to the definition of an ellipse,the sum of distance

Vedclass · Accepted Answer

$\frac{x^2}{a^2} + \frac{y^2}{a^2(1 - e^2)} = 1$

Vedclass · Answer

(A) Let $P(x, y)$ be the moving point and $F_1(ae, 0)$,$F_2(-ae, 0)$ be the fixed points.According to the definition of an ellipse,the sum of distances from two fixed points (foci) is constant $(2a)$.$PF_1 + PF_2 = 2a$$\sqrt{(x - ae)^2 + y^2} + \sqrt{(x + ae)^2 + y^2} = 2a$$\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}$$a - ex = \sqrt{(x - ae)^2 + y^2}$Squaring again:$a^2 - 2aex + e^2x^2 = x^2 - 2aex + a^2e^2 + y^2$$x^2(1 - e^2) + y^2 = a^2(1 - e^2)$Dividing by $a^2(1 - e^2)$:$\frac{x^2}{a^2} + \frac{y^2}{a^2(1 - e^2)} = 1$.

$A$ point moves such that the sum of its distances from two fixed points $(ae, 0)$ and $(-ae, 0)$ is always $2a$. Then the equation of its locus is

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