Question

The coordinates of the foci of the hyperbola xy = c² are

• A. (±c, ± c)
• B. ($\pm \frac{c}{\sqrt{2}}$, $\pm \frac{c}{\sqrt{2}})$
• C. (±2c, ±2c)
• D. ($\pm \sqrt{2c}$, $\pm \sqrt{2c}$)

Answer 1

Answer: (D)

($\pm \sqrt{2c}$, $\pm \sqrt{2c}$)

Answer 2

The given equation of the hyperbola is $xy = c^2$. This is a rectangular hyperbola whose asymptotes are the coordinate axes.

To find the foci, we can rotate the coordinate axes by an angle of $45^\circ$. Let the new coordinates be $(X, Y)$. The transformation equations are:

$x = X \cos 45^\circ - Y \sin 45^\circ = X \frac{1}{\sqrt{2}} - Y \frac{1}{\sqrt{2}} = \frac{X-Y}{\sqrt{2}}$

$y = X \sin 45^\circ + Y \cos 45^\circ = X \frac{1}{\sqrt{2}} + Y \frac{1}{\sqrt{2}} = \frac{X+Y}{\sqrt{2}}$

Substitute these into the equation $xy = c^2$:

$(\frac{X-Y}{\sqrt{2}})(\frac{X+Y}{\sqrt{2}}) = c^2$

$\frac{X^2 - Y^2}{2} = c^2$

$X^2 - Y^2 = 2c^2$

$\frac{X^2}{2c^2} - \frac{Y^2}{2c^2} = 1$

This is the standard form of a hyperbola $\frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1$, where $a^2 = 2c^2$ and $b^2 = 2c^2$.

For this hyperbola in the $(X, Y)$ system:

The semi-major axis is $a = \sqrt{2c^2} = \sqrt{2}|c|$. We can assume $c > 0$ without loss of generality for finding coordinates, the $\pm$ signs will handle the case $c < 0$. So $a = \sqrt{2}c$.

The semi-minor axis is $b = \sqrt{2c^2} = \sqrt{2}c$.

The eccentricity $e$ is given by $e^2 = 1 + \frac{b^2}{a^2}$.

$e^2 = 1 + \frac{2c^2}{2c^2} = 1 + 1 = 2$

$e = \sqrt{2}$ (since eccentricity is always positive).

The foci of the hyperbola $\frac{X^2}{a^2} - \frac{Y^2}{b^2} = 1$ in the $(X, Y)$ system are located at $(\pm ae, 0)$.

Foci in $(X, Y)$ are $(\pm (\sqrt{2}c)(\sqrt{2}), 0) = (\pm 2c, 0)$.

The two foci are $(2c, 0)$ and $(-2c, 0)$ in the $(X, Y)$ system.

Now we need to convert these coordinates back to the original $(x, y)$ system using the inverse transformation equations:

$x = \frac{X-Y}{\sqrt{2}}$

$y = \frac{X+Y}{\sqrt{2}}$

For the focus $(X, Y) = (2c, 0)$:

$x = \frac{2c - 0}{\sqrt{2}} = \frac{2c}{\sqrt{2}} = \sqrt{2}c$

$y = \frac{2c + 0}{\sqrt{2}} = \frac{2c}{\sqrt{2}} = \sqrt{2}c$

So, one focus is $(\sqrt{2}c, \sqrt{2}c)$.

For the focus $(X, Y) = (-2c, 0)$:

$x = \frac{-2c - 0}{\sqrt{2}} = \frac{-2c}{\sqrt{2}} = -\sqrt{2}c$

$y = \frac{-2c + 0}{\sqrt{2}} = \frac{-2c}{\sqrt{2}} = -\sqrt{2}c$

So, the other focus is $(-\sqrt{2}c, -\sqrt{2}c)$.

The coordinates of the foci are $(\sqrt{2}c, \sqrt{2}c)$ and $(-\sqrt{2}c, -\sqrt{2}c)$.

This set of points can be represented as $(\pm \sqrt{2}c, \pm \sqrt{2}c)$, where the signs are taken together (i.e., $(+,+)$ and $(-,-)$).