Reflect the hyperbola given by the equation $x y=1$ over the line $y=2 x$. Write the equation of the reflection.
I. solution. The hyperbola given by the equation $x \cdot y=1$ is a so-called "isosceles" hyperbola, whose axes are equal. Its real axis is the segment between the points $(1 ; 1)$ and $(-1 ;-1)$, whose length is: $2 a=2 \sqrt{2}$. Therefore, the imaginary axis is: $2 b=2 \sqrt{2}$. It is known that for every hyperbola, $a^{2}+b^{2}=c^{2}$, where $c$ is half the distance between the foci, so $c=2$. Therefore, the two foci are: $F_{1}(\sqrt{2} ; \sqrt{2})$, and $F_{2}(-\sqrt{2} ;-\sqrt{2})$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e1e319d395a9169fac79g-1.jpg?height=672&width=637&top_left_y=355&top_left_x=733)

The lengths of the axes of the reflected hyperbola are obviously unchanged, and let the foci be $G_{1}$ and $G_{2}$. Reflecting $F_{1}$ and $F_{2}$ over the line $y=2 x$, the coordinates of the points $G_{1}$ and $G_{2}$ can be calculated as follows:

$$
G_{1}\left(\frac{\sqrt{2}}{5} ; \frac{7 \sqrt{2}}{5}\right) \text { and } G_{2}\left(-\frac{\sqrt{2}}{5} ;-\frac{7 \sqrt{2}}{5}\right)
$$

By the definition of a hyperbola, the hyperbola with foci $G_{1}$ and $G_{2}$ and real axis $2 a=2 \sqrt{2}$ consists of all points $\mathrm{P}(\mathrm{x} ; \mathrm{y})$ such that:

$$
G_{1} P-G_{2} P=2 \sqrt{2} \text { or } G_{2} P-G_{1} P=2 \sqrt{2}
$$

Therefore, the equation of the hyperbola is:

$$
\left|\sqrt{\left(x-\frac{\sqrt{2}}{5}\right)^{2}+\left(y-\frac{7 \sqrt{2}}{5}\right)^{2}}-\sqrt{\left(x+\frac{\sqrt{2}}{5}\right)^{2}+\left(y+\frac{7 \sqrt{2}}{5}\right)^{2}}\right|=2 \sqrt{2}
$$

Squaring and rearranging:

$$
x^{2}+y^{2}=\sqrt{\left(x-\frac{\sqrt{2}}{5}\right)^{2}+\left(y-\frac{7 \sqrt{2}}{5}\right)^{2}} \cdot \sqrt{\left(x+\frac{\sqrt{2}}{5}\right)^{2}+\left(y+\frac{7 \sqrt{2}}{5}\right)^{2}}
$$

Equation (3) is a consequence of the two equations connected by "or" in (1) (or (2)), but (3) also implies the two equations in (1). Therefore, (1), (2), and (3) are pairwise equivalent. Since both sides of (3) are non-negative, squaring them results in an equivalent equation:

$$
\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}+y^{2}+4-\frac{2 x \sqrt{2}}{5}-\frac{14 y \sqrt{2}}{5}\right) \cdot\left(x^{2}+y^{2}+4+\frac{2 x \sqrt{2}}{5}+\frac{14 y \sqrt{2}}{5}\right)
$$

From this,

$$
\left(x^{2}+y^{2}\right)^{2}=\left(x^{2}+y^{2}+4\right)^{2}-\left(\frac{2 x \sqrt{2}}{5}+\frac{14 y \sqrt{2}}{5}\right)^{2}
$$

which simplifies to

$$
8 x^{2}+8 y^{2}+16=\frac{8 x^{2}+2 \cdot 56 x y+8 \cdot 49 y^{2}}{25}
$$

Finally, after rearranging, the equation of the reflected hyperbola is:

$$
12 y^{2}+7 x y-12 x^{2}=25
$$

II. solution. Let $P_{0}\left(x_{0} ; y_{0}\right)$ be an arbitrary point on the hyperbola given by the equation $x \cdot y=1$, and let its reflection over the line $y=2 x$ be $P(x ; y)$. Since the line $P_{0} P$ is perpendicular to $y=2 x$, its slope is $-1 / 2$, and thus

$$
\frac{y-y_{0}}{x-x_{0}}=-\frac{1}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_e1e319d395a9169fac79g-2.jpg?height=493&width=471&top_left_y=529&top_left_x=816)

The midpoint of the segment $P_{0} P$, $\left(\frac{x_{0}+x}{2} ; \frac{y_{0}+y}{2}\right)$, lies on the line $y=2 x$, so

$$
\frac{y+y_{0}}{2}=2 \cdot \frac{x+x_{0}}{2}
$$

Solving the system of equations (4) and (5), the original coordinates in terms of the new ones are:

$$
x_{0}=\frac{4 y-3 x}{5} ; \quad y_{0}=\frac{4 x+3 y}{5}
$$

Since $x_{0} \cdot y_{0}=1$, the equation of the reflected hyperbola is

$$
\frac{4 y-3 x}{5} \cdot \frac{4 x+3 y}{5}=1
$$

which simplifies to

$$
12 y^{2}+7 x y-12 x^{2}=25
$$