Let $E$ be the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$. For any three distinct points $P,Q$ and $Q′$ on $E$, let $M(P, Q)$ be the mid-point of the line segment joining $P$ and $Q$, and $M(P, Q′)$ be the mid-point of the line segment joining $P$ and $Q′$. Then the maximum possible value of the distance between $M(P,Q)$ and $M(P,Q′)$, as $P,Q$ and $Q ′$ vary on $E$, is ___ .
My approach is as follow
Let $P(4cos\alpha,3sin\alpha)$,$Q(4cos\theta_1,3sin\theta_1)$,$Q′(4cos\theta_2,3sin\theta_2)$
$M(P, Q)=(\frac{4(cos\theta_1+cos\theta_2)}{2},)$
Let the mid point be represented as $M\left( {P,Q} \right) = \left( {\frac{{4\left( {\cos \alpha + \cos {\theta _1}} \right)}}{2},\frac{{3\left( {\sin \alpha + \sin {\theta _1}} \right)}}{2}} \right);M\left( {P,Q'} \right) = \left( {\frac{{4\left( {\cos \alpha + \cos {\theta _2}} \right)}}{2},\frac{{3\left( {\sin \alpha + \sin {\theta _2}} \right)}}{2}} \right)$
$\frac{1}{4}\sqrt {{{\left( {4\cos \alpha + 4\cos {\theta _1} – \left( {4\cos \alpha + 4\cos {\theta _2}} \right)} \right)}^2} + {{\left( {3\sin \alpha + 3\sin {\theta _1} – \left( {3\sin \alpha + 3\sin {\theta _2}} \right)} \right)}^2}} $
$\frac{1}{4}\sqrt {16\left( {{{\cos }^2}{\theta _1} + {{\cos }^2}{\theta _2} – 2\cos {\theta _1}\cos {\theta _2}} \right) + 9\left( {{{\sin }^2}{\theta _1} + {{\sin }^2}{\theta _2} – 2\sin {\theta _1}\sin {\theta _2}} \right)} $
$\frac{1}{4}\sqrt {16{{\cos }^2}{\theta _1} + 9{{\sin }^2}{\theta _1} + 16{{\cos }^2}{\theta _2} + 9{{\sin }^2}{\theta _2} – 32\cos {\theta _1}\cos {\theta _2} – 18\sin {\theta _1}\sin {\theta _2}} $
$\frac{1}{4}\sqrt {18 + 7\left( {{{\cos }^2}{\theta _1} + {{\cos }^2}{\theta _2}} \right) – 14\cos {\theta _1}\cos {\theta _2} – 18\left( {\cos \left( {{\theta _1} – {\theta _2}} \right)} \right)} $
${\cos ^2}{\theta _1} = \frac{{1 + \cos 2{\theta _1}}}{2};{\cos ^2}{\theta _2} = \frac{{1 + \cos 2{\theta _2}}}{2}$
${\cos ^2}{\theta _1} + {\cos ^2}{\theta _2} = 1 + \frac{{2\cos \left( {{\theta _1} + {\theta _2}} \right)\cos \left( {{\theta _1} – {\theta _2}} \right)}}{2} = 1 + \cos \left( {{\theta _1} + {\theta _2}} \right)\cos \left( {{\theta _1} – {\theta _2}} \right)$
$\frac{1}{4}\sqrt {18 + 7\left( {1 + \cos \left( {{\theta _1} + {\theta _2}} \right)\cos \left( {{\theta _1} – {\theta _2}} \right)} \right) – 7\left( {\cos \left( {{\theta _1} + {\theta _2}} \right) + \cos \left( {{\theta _1} – {\theta _2}} \right)} \right) – 18\left( {\cos \left( {{\theta _1} – {\theta _2}} \right)} \right)} $
$\frac{1}{4}\sqrt {25 + 7\cos \left( {{\theta _1} + {\theta _2}} \right)\left( {\cos \left( {{\theta _1} – {\theta _2}} \right) – 1} \right) – 25\left( {\cos \left( {{\theta _1} – {\theta _2}} \right)} \right)} $
How do we proceed from here , we get the official answer 4 when $\theta_1=0$ & $\theta_2=\pi$
Best Answer
Define $X\equiv M(P,Q)$ and $Y\equiv M(P,Q')$. Using the midpoint theorem $(\triangle PXY\sim \triangle PQQ')$, $$|XY|=\frac{1}{2}\cdot|QQ'|$$ $|QQ'|$ is maximum when $Q$ and $Q'$ lie on the major axis of the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}$ (the two farthest points on an ellipse). Hence, $Q=(\pm 4,0)$ and $Q'=(\mp4,0)$.
Therefore, $$\max(|XY|)=\frac{1}{2}\cdot8=\boxed{4}$$