A point P moves such that the sum of its dist | vedclass

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Question

(A) Let $P(h, k)$ be the moving point such that the sum of its distances from $A(ae, 0)$ and $B(-ae, 0)$ is $2a$.$PA + PB = 2a$$\sqrt{(h - ae)^2 + k^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $P(h, k)$ be the moving point such that the sum of its distances from $A(ae, 0)$ and $B(-ae, 0)$ is $2a$.$PA + PB = 2a$$\sqrt{(h - ae)^2 + k^2} + \sqrt{(h + ae)^2 + k^2} = 2a$$\sqrt{(h - ae)^2 + k^2} = 2a - \sqrt{(h + ae)^2 + k^2}$Squaring both sides:$(h - ae)^2 + k^2 = 4a^2 + (h + ae)^2 + k^2 - 4a\sqrt{(h + ae)^2 + k^2}$$h^2 - 2aeh + a^2e^2 + k^2 = 4a^2 + h^2 + 2aeh + a^2e^2 + k^2 - 4a\sqrt{(h + ae)^2 + k^2}$$-4aeh - 4a^2 = -4a\sqrt{(h + ae)^2 + k^2}$$eh + a = \sqrt{(h + ae)^2 + k^2}$Squaring again:$e^2h^2 + 2aeh + a^2 = h^2 + 2aeh + a^2e^2 + k^2$$h^2(1 - e^2) + k^2 = a^2(1 - e^2)$Dividing by $a^2(1 - e^2)$:$\frac{h^2}{a^2} + \frac{k^2}{a^2(1 - e^2)} = 1$Thus,the locus of $(h, k)$ is $\frac{x^2}{a^2} + \frac{y^2}{a^2(1 - e^2)} = 1$.

$A$ point $P$ moves such that the sum of its distances from the points $(ae, 0)$ and $(-ae, 0)$ is always $2a$. Find the locus of $P$ (where $0 < e < 1$).

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