Q.The area of the region bounded by the ellipse $\frac{x^2}{25} + \frac{y^2}{16} = 1$ is
(A) $20\pi$ sq units
(B) $20\pi^2$ sq units
(C) $16\pi^2$ sq units
(D) $25\pi$ sq units

The area of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\pi a b$.

Here $a^2 = 25 \Rightarrow a = 5$, and $b^2 = 16 \Rightarrow b = 4$. …

The area of an ellipse is one of those results that feels like magic until you see why it works. The formula $\pi a b$ is a natural generalization of the area of a circle ($\pi r^2$). Think of a circle as a special ellipse where $a = b = r$. When you stretch a circle along one axis, the area scales proportionally — that’s the intuition behind the formula.

But let’s derive it properly for this specific ellipse, so you never have to memorize blindly.

1. Identify the semi-axes.

   The given ellipse is $\frac{x^2}{25} + \frac{y^2}{16} = 1$.

   Compare with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

   Here $a^2 = 25$, so $a = 5$ (the semi-major axis along $x$).

   And $b^2 = 16$, so $b = 4$ (the semi-minor axis along $y$).

2. Set up the area integral.

   The ellipse is symmetric about both axes. So we can find the area in the first quadrant and multiply by 4.

   From the equation, solve for $y$ in the first quadrant:

The area of the whole ellipse is

3. Evaluate the integral. The integral $\int \sqrt{a^2 - x^2} \, dx$ is a standard form. For $a = 5$:

(This is the area of a quarter-circle of radius 5 — a neat shortcut.) …

Method: Area enclosed by a standard ellipse

Use this whenever the region is the full interior of an ellipse written in standard form $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$.

Steps

Step 1: Read $a^2$ and $b^2$ off the denominators.

The number under $x^2$ is $a^2$; the number under $y^2$ is $b^2$. Take square roots to get the semi-axes $a$ and $b$ — this is the single most error-prone step, so do it explicitly.

Step 2: Apply the area formula.

$\text{Area} = \pi\, a\, b$ …

Common Mistakes

Mistake 1: Treating the denominators as $a$ and $b$ directly.

Why it's wrong: the standard form has $a^2$ under $x^2$ and $b^2$ under $y^2$, so here $a^2 = 25,\ b^2 = 16$ give $a = 5,\ b = 4$ — not $25$ and $16$. Using $25 \times 16 \times \pi$ is badly wrong. Correct approach: take square roots first, then $\text{Area} = \pi a b = 20\pi$.

Mistake 2: Adding a stray $\pi^2$. …