An ellipse with its centre at the origin and its axes along the coordinate axes passes through the points (2, 2) and (4, 1). The area (in sq. units) of the region enclosed by the ellipse is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(10\pi\)
Let the ellipse be \(\displaystyle\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\), and write \(\displaystyle p = \frac{1}{a^{2}}\), \(\displaystyle q = \frac{1}{b^{2}}\). The two points give
(2, 2): \(4p+4q = 1\)
(4, 1): \(16p+q = 1\)
From the second equation q = 1 − 16p. Putting this in the first:
\(4p+4-64p = 1\)
\(60p = 3\)
\( \displaystyle p = \frac{1}{20},\ \ q = 1-\frac{16}{20} = \frac{1}{5} \)
So \(a^{2} = 20\) and \(b^{2} = 5\), that is \(a = 2\sqrt{5}\) and \(b = \sqrt{5}\).
The area enclosed by an ellipse is \(\pi ab\):
\(A = \pi\times2\sqrt{5}\times\sqrt{5} = 10\pi\)
Why the other options are incorrect
Option (1) \(20\pi\) — Incorrect. This is \(\pi a^{2}\), the area of the circle on the major axis, not of the ellipse.
Option (2) \(5\pi\) — Incorrect. This is \(\pi b^{2}\), the area of the circle on the minor axis.
Option (4) \(100\pi\) — Incorrect. This multiplies \(a^{2}\) and \(b^{2}\); the area uses a and b, their square roots.