Find the area of the greatest rectangle that can be inscribed in the ellipse .

Consider the rectangle ABCD having coordinates as A(-a, b), B(a, b), C(a, -b) and D(-a, -b)

Hence from figure the length CD and breadth BC are 2a and 2b respectively

Let A be the area of rectangle

⇒ A = length × breadth

⇒ A = 2a × 2b

⇒ A = 4ab

We need to maximise area A based on length or breadth so we will require area either only in terms of length a or breadth b

(a, b) lies on ellipse which means it will satisfy the equation

Substitute value of a in A = 4ab

Differentiate A with respect to b

Using uv rule which states that (uv)’ = u’v + uv’

Here u = b and

Now we will find the critical points where

⇒ k² – 2b² = 0

⇒ (k + √2b)(k – √2b) = 0

We don’t need to check whether it’s a point of minima or maxima because the minimum area of rectangle will be 0 which we will get when b = 0 but here we have got b as non-zero hence we can skip the verification part

As now we have got b we will now find a by putting b in ellipse equation (i)

As we are squaring b hence it doesn’t matter whether b is positive or negative hence taking b as

Put values of a and b in A = 4ab from (p) and (q)

So the answer will be positive or negative but area cannot be negative hence consider only positive part

⇒ A = 2hk

Hence the maximum area of rectangle in ellipse will be 2hk sq. units