Match List-I with List-II
| List-I (Region) | List-II (Area in sq. units) | ||
| (A) | \(\left\{(x, y): \displaystyle\frac{x^{2}}{9}+\frac{y^{2}}{4}\le 1\right\}\) | (I) | \(9\pi\) |
| (B) | \(\left\{(x, y): \displaystyle\frac{x^{2}}{25}+\frac{y^{2}}{16}\le 1,\ y \le 0\right\}\) | (II) | \(2\pi\) |
| (C) | \(\left\{(x, y): \displaystyle\frac{x^{2}}{4}+\frac{y^{2}}{16}\le 1,\ x \ge 0,\ y \ge 0\right\}\) | (III) | \(6\pi\) |
| (D) | \(\left\{(x, y): \displaystyle\frac{x^{2}}{36}+\frac{y^{2}}{9}\le 1,\ x \le 0\right\}\) | (IV) | \(10\pi\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
The ellipse \(\displaystyle\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\) encloses the area \(\pi ab\). It is symmetric about both axes, so each axis cuts it into two equal halves, and each quadrant holds one quarter.
(A) \(\left\{(x, y): \displaystyle\frac{x^{2}}{9}+\frac{y^{2}}{4}\le 1\right\}\)
This matches (III) \(6\pi\).
Whole ellipse with a = 3, b = 2: \(\pi\times3\times2 = 6\pi\)
(B) \(\left\{(x, y): \displaystyle\frac{x^{2}}{25}+\frac{y^{2}}{16}\le 1,\ y \le 0\right\}\)
This matches (IV) \(10\pi\).
Lower half with a = 5, b = 4: \(\displaystyle\frac{1}{2}\times\pi\times5\times4 = 10\pi\)
(C) \(\left\{(x, y): \displaystyle\frac{x^{2}}{4}+\frac{y^{2}}{16}\le 1,\ x \ge 0,\ y \ge 0\right\}\)
This matches (II) \(2\pi\).
First-quadrant quarter with a = 2, b = 4: \(\displaystyle\frac{1}{4}\times\pi\times2\times4 = 2\pi\)
(D) \(\left\{(x, y): \displaystyle\frac{x^{2}}{36}+\frac{y^{2}}{9}\le 1,\ x \le 0\right\}\)
This matches (I) \(9\pi\).
Left half with a = 6, b = 3: \(\displaystyle\frac{1}{2}\times\pi\times6\times3 = 9\pi\)