Ellipse Question - None

Question

Area of the greatest rectangle that can be inscribed in the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is

Accepted Answer

Step 1: Let a rectangle inscribed in the ellipse have vertices at (±x₀, ±y₀) where (x₀, y₀) is on the ellipse.

Step 2: The area of the rectangle is A = 4x₀y₀

Step 3: Since (x₀, y₀) lies on the ellipse: $\frac{x_0^2}{a^2} + \frac{y_0^2}{b^2} = 1$, so $y_0 = b\sqrt{1 - \frac{x_0^2}{a^2}}$

Step 4: Substitute into area: $A = 4x_0 \cdot b\sqrt{1 - \frac{x_0^2}{a^2}}$

Step 5: To maximize, let $u = \frac{x_0^2}{a^2}$. Then $A = 4ab\sqrt{u(1-u)}$

Step 6: The function $f(u) = \sqrt{u(1-u)}$ is maximized when $\frac{d}{du}[u(1-u)] = 0$, giving $1 - 2u = 0$, so $u = \frac{1}{2}$

Step 7: When $u = \frac{1}{2}$: $\sqrt{\frac{1}{2} \cdot \frac{1}{2}} = \frac{1}{2}$

Step 8: Maximum area = $4ab \cdot \frac{1}{2} = 2ab$

∴ Answer: 2ab

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