Consider the ellipse \(E: \displaystyle\frac{x^{2}}{16}+\frac{y^{2}}{4}=1\) and the circle \(C: x^{2}+y^{2}=16\). Which of the following statements are correct?
(A) E and C meet only at the points (4, 0) and (−4, 0)
(B) The area of the region enclosed by E is \(8\pi\) sq. units
(C) The area of the region inside C and outside E is \(8\pi\) sq. units
(D) The area of the region enclosed by E is one-fourth of the area of the region enclosed by C
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (C) only
From E, \(x^{2} = 16-4y^{2}\). Putting this in C:
\(16-4y^{2}+y^{2} = 16\)
\(y = 0,\ \ x = \pm4\)
At any point of E, \(x^{2}+y^{2} = 16-3y^{2} \le 16\), so E lies inside C, touching it only at the ends of the major axis.
The ellipse has semi-axes 4 and 2, and the circle has radius 4:
Area of E \(= \pi\times4\times2 = 8\pi\)
Area of C \(= \pi\times4^{2} = 16\pi\)
Area inside C and outside E \(= 16\pi-8\pi = 8\pi\)
So E and C meet only at (±4, 0), E encloses \(8\pi\), C encloses \(16\pi\), and the region between them has area \(8\pi\).
Checking the statements
(A) E and C meet only at the points (4, 0) and (−4, 0) — Correct. Solving the two equations together gives y = 0, x = ±4.
(B) The area of the region enclosed by E is \(8\pi\) sq. units — Correct. \(\pi\times4\times2 = 8\pi\).
(C) The area of the region inside C and outside E is \(8\pi\) sq. units — Correct. \(16\pi-8\pi = 8\pi\).
(D) The area of the region enclosed by E is one-fourth of the area of the region enclosed by C — Incorrect. \(8\pi\) is one-half of \(16\pi\).