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 $\mathrm{M}\left(\mathrm{P}, \mathrm{Q}^{\prime}\right)$ 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 $\mathrm{P}, \mathrm{Q}$ and $\mathrm{Q}^{\prime}$ vary on E , is ____.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Ellipse

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 ____.

Answer: 4

Solution

Let A = M(P, Q) and B = M(P, Q'). Then A is the midpoint of PQ and B is the midpoint of PQ'.

By the midpoint theorem in triangle PQQ', the line segment AB is parallel to QQ' and

A B = \frac{1}{2} ∣ Q Q^{'} ∣

.

The maximum value of

∣ Q Q^{'} ∣

occurs when Q and Q' are at opposite ends of the major axis of the ellipse, giving

∣ Q Q^{'} ∣_{m a x} = 2 a = 2 (4) = 8

.

Therefore, the maximum value of AB is

\frac{8}{2} = 4

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Integer answer

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JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Ellipse

Solution

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