Question

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is such that it has least area but contains the circle $(x-1)^2+y^2=1$. If eccentricity of ellipse is e, then value of $3e^2$ is equal to

• A. 3/2
• B. 1/2
• C. 2/3
• D. 3

Answer 1

Answer: (A)

3/2

Answer 2

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ contains the circle $(x-1)^2+y^2=1$. The area of the ellipse is $A = \pi ab$. To minimize the area, the ellipse must be tangent to the circle.

The points on the circle can be parameterized as $(1+\cos\theta, \sin\theta)$. For the ellipse to contain the circle, all these points must satisfy the ellipse equation: $\frac{(1+\cos\theta)^2}{a^2} + \frac{\sin^2\theta}{b^2} \le 1 \quad \forall \theta$ The maximum value of the left side must be equal to 1 for the ellipse to have the least area while containing the circle. Let $f(\theta) = \frac{(1+\cos\theta)^2}{a^2} + \frac{\sin^2\theta}{b^2}$. The maximum of $f(\theta)$ occurs when $\cos\theta = \frac{b^2}{a^2-b^2}$ (assuming $a>b$) or at $\theta=0$. At $\theta=0$, $f(0) = \frac{(1+1)^2}{a^2} + \frac{0^2}{b^2} = \frac{4}{a^2}$. So $\frac{4}{a^2} \le 1 \implies a \ge 2$. When $a>b$, the maximum value of $f(\theta)$ is $\frac{a^2}{b^2(a^2-b^2)}$, which must be equal to 1. $\frac{a^2}{b^2(a^2-b^2)} = 1 \implies a^2 = a^2b^2 - b^4 \implies b^4 - a^2b^2 + a^2 = 0$ To minimize the area $\pi ab$, we need to minimize $a^2b^2$. From the quadratic equation for $b^2$, $b^2 = \frac{a^2 \pm \sqrt{a^4-4a^2}}{2}$. We take the minus sign for minimum $b^2$: $b^2 = \frac{a^2 - \sqrt{a^4-4a^2}}{2}$. The minimum area occurs when $a^4-4a^2=0$, which implies $a^2=4$, so $a=2$. Then $b^2 = \frac{4 - 0}{2} = 2$, so $b=\sqrt{2}$. The semi-axes are $a=2$ and $b=\sqrt{2}$. Since $a>b$, $a$ is the semi-major axis. The eccentricity $e$ is given by $b^2 = a^2(1-e^2)$. $2 = 4(1-e^2) \implies \frac{1}{2} = 1-e^2 \implies e^2 = \frac{1}{2}$. The value of $3e^2$ is $3 \times \frac{1}{2} = \frac{3}{2}$.