Question

Consider three curves

$C_1: x^2 - y^2 = 1 (x < 0)$

$C_2: xy = 1 (y > 0)$

$C_3$: Ellipse whose one focus is also focus of $C_2$ and corresponding directrix of this focus is asymptote of curve $C_1$.

Also length of major axis of $C_3$ is $\frac{8}{3}$.

On the basis of above information, answer the following questions: Eccentricity of $C_3$ is equal to -

• A. $\frac{1}{4}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{2}$

Answer 1

Answer: (D)

$\frac{1}{2}$

Answer 2

The problem involves three curves $C_1, C_2, C_3$.

$C_1: x^2 - y^2 = 1 (x < 0)$. This is a hyperbola. The asymptotes are $y = \pm x$.

$C_2: xy = 1 (y > 0)$. This is a rectangular hyperbola. The foci are $(\sqrt{2}, \sqrt{2})$ and $(-\sqrt{2}, -\sqrt{2})$. The corresponding directrices are $x+y=\sqrt{2}$ and $x+y=-\sqrt{2}$.

$C_3$: Ellipse. One focus $F_3$ is a focus of $C_2$, and the corresponding directrix $d_3$ is an asymptote of $C_1$.

Let $F_3$ be a focus of $C_2$. So $F_3 = (\sqrt{2}, \sqrt{2})$ or $(-\sqrt{2}, -\sqrt{2})$.

Let $d_3$ be an asymptote of $C_1$. So $d_3$ is $y=x$ (i.e., $x-y=0$) or $y=-x$ (i.e., $x+y=0$).

The distance from a focus to its corresponding directrix is $d$. For an ellipse with semi-major axis $a$ and eccentricity $e$, this distance is $d = \frac{a}{e} - ae = a(\frac{1}{e}-e)$.

The distance from $F_3$ to $d_3$ must be non-zero, as a focus cannot lie on its directrix.

If $F_3 = (\sqrt{2}, \sqrt{2})$ and $d_3: x-y=0$, distance is $\frac{|\sqrt{2}-\sqrt{2}|}{\sqrt{1^2+(-1)^2}} = 0$. Not possible.

If $F_3 = (\sqrt{2}, \sqrt{2})$ and $d_3: x+y=0$, distance is $\frac{|\sqrt{2}+\sqrt{2}|}{\sqrt{1^2+1^2}} = \frac{2\sqrt{2}}{\sqrt{2}} = 2$. Possible.

If $F_3 = (-\sqrt{2}, -\sqrt{2})$ and $d_3: x-y=0$, distance is $\frac{|-\sqrt{2}-(-\sqrt{2})|}{\sqrt{1^2+(-1)^2}} = 0$. Not possible.

If $F_3 = (-\sqrt{2}, -\sqrt{2})$ and $d_3: x+y=0$, distance is $\frac{|-\sqrt{2}+(-\sqrt{2})|}{\sqrt{1^2+1^2}} = \frac{|-2\sqrt{2}|}{\sqrt{2}} = 2$. Possible.

So, the distance from the focus to the directrix is $d=2$.

The length of the major axis of $C_3$ is $2a = \frac{8}{3}$, so $a = \frac{4}{3}$.

Using the relation $d = a(\frac{1}{e}-e)$, we have $2 = \frac{4}{3}(\frac{1}{e}-e)$.

$\frac{3}{2} = \frac{1}{e}-e = \frac{1-e^2}{e}$.

$3e = 2(1-e^2) = 2 - 2e^2$.

$2e^2 + 3e - 2 = 0$.

This is a quadratic equation in $e$. Using the quadratic formula:

$e = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm \sqrt{25}}{4} = \frac{-3 \pm 5}{4}$.

Since eccentricity $e$ must be positive for an ellipse ($0 < e < 1$), we take the positive root:

$e = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$.

The eccentricity of $C_3$ is $\frac{1}{2}$.