2.2.1. Derive the equation of the line for which the ratio of the distance from any point to the point $F(2 ; 4)$ to the distance from the same point to the line $l: x=-4$ is 2. Reduce the equation of the line to its canonical form and determine the type of this curve.
Solution. Let $M(x ; y)$ be the current point on the line. From point $M$, we drop a perpendicular to the line $x=-4$, which intersects it at point $N(-4 ; y)$. According to the problem's condition: $\frac{|M F|}{|M N|}=2$, or $|M F|=2|M N|$.

Then

$$
\sqrt{\left(x_{M}-x_{F}\right)^{2}+\left(y_{M}-y_{F}\right)^{2}}=2 \sqrt{\left(x_{M}-x_{N}\right)^{2}+\left(y_{M}-y_{N}\right)^{2}}
$$

or

$$
\sqrt{(x-2)^{2}+(y-4)^{2}}=2 \sqrt{(x+4)^{2}+(y-y)^{2}} .
$$

Squaring both sides, expanding the brackets, and combining like terms, we get:

$$
3 x^{2}+36 x+60-(y-4)^{2}=0
$$

We make the coefficient of $x^{2}$ equal to one by dividing the entire equation by 3:

$$
x^{2}+12 x+20-\frac{1}{3}(y-4)^{2}=0 .
$$

We write the polynomial depending on $x$ as a complete square:

$x^{2}+12 x=x^{2}+2 \cdot x \cdot 6=x^{2}+2 \cdot x \cdot 6+36-36=(x+6)^{2}-36$.

Then the equation becomes:

$$
(x+6)^{2}-36+20-\frac{1}{3}(y-4)^{2}=0,
$$

further

$$
(x+6)^{2}-\frac{1}{3}(y-4)^{2}=16
$$

or, dividing by 16, we have:

$$
\frac{(x+6)^{2}}{16}-\frac{(y-4)^{2}}{48}=1
$$

Introducing a new coordinate system:

$$
\begin{aligned}
& X=x+6 \\
& Y=y-4,
\end{aligned}
$$

we bring the equation of the line to its canonical form:

$$
\frac{X^{2}}{4^{2}}-\frac{Y^{2}}{(4 \sqrt{3})^{2}}=1
$$

This is the canonical equation of a hyperbola.

Remark. If $\frac{m}{n}=1$, we arrive at the canonical equation of a parabola $Y^{2}=2 p X$, and if $\frac{m}{n}<1$, we obtain the canonical equation of an ellipse:

$$
\frac{X^{2}}{a^{2}}+\frac{Y^{2}}{b^{2}}=1
$$

## 2.3. Plane and Line in Space