Question

Let E be the ellipse $\frac{x^2}{25} + \frac{y^2}{4} = 1$. For any three distinct points A, B, C on the ellipse E, let $M_1$ be the mid-point of the segment joining points A and B and $M_2$ be the mid-point of the segment joining points A and C. Then the maximum possible value of the distance between $M_1$ and $M_2$ as A, B, C vary on the ellipse E, is ____.

Answer 1

5

Answer 2

Let the equation of the ellipse be $E: \frac{x^2}{25} + \frac{y^2}{4} = 1$. From the equation, $a = 5$ and $b = 2$.

Let $A = (5 \cos \alpha, 2 \sin \alpha)$, $B = (5 \cos \beta, 2 \sin \beta)$, and $C = (5 \cos \gamma, 2 \sin \gamma)$. Then $M_1 = \left( \frac{5 \cos \alpha + 5 \cos \beta}{2}, \frac{2 \sin \alpha + 2 \sin \beta}{2} \right) = \left( \frac{5}{2}(\cos \alpha + \cos \beta), \sin \alpha + \sin \beta \right)$ $M_2 = \left( \frac{5 \cos \alpha + 5 \cos \gamma}{2}, \frac{2 \sin \alpha + 2 \sin \gamma}{2} \right) = \left( \frac{5}{2}(\cos \alpha + \cos \gamma), \sin \alpha + \sin \gamma \right)$ Then \begin{align*} M_1 M_2 &= \sqrt{\left( \frac{5}{2}(\cos \gamma - \cos \beta) \right)^2 + (\sin \gamma - \sin \beta)^2} \ &= \sqrt{\frac{25}{4} (\cos \gamma - \cos \beta)^2 + (\sin \gamma - \sin \beta)^2} \ &= \frac{1}{2} \sqrt{25 (\cos \gamma - \cos \beta)^2 + 4 (\sin \gamma - \sin \beta)^2} \end{align*} The distance between $B$ and $C$ is $BC = \sqrt{(5 \cos \gamma - 5 \cos \beta)^2 + (2 \sin \gamma - 2 \sin \beta)^2} = \sqrt{25 (\cos \gamma - \cos \beta)^2 + 4 (\sin \gamma - \sin \beta)^2}$ Therefore, $M_1 M_2 = \frac{1}{2} BC$. The maximum distance between two points on the ellipse is $2a = 10$. Therefore, the maximum possible value of $M_1 M_2$ is $\frac{1}{2} (10) = 5$.