Example 6. On the hyperbola with the equation $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$, find the point $M$ such that $M F_{1}=2 M F_{2}$. Write the equations of the asymptotes and the directrices of the hyperbola. Find its eccentricity. Make a sketch.
Solution. 1) We have $a=4, b=3, c^{2}=a^{2}+b^{2}, c=5$. We construct the hyperbola as follows (Fig. 2.22): in a rectangle with sides $x= \pm a$ and $y= \pm b$

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-029.jpg?height=399&width=598&top_left_y=470&top_left_x=80)

Fig. 2.22 (i.e., $x= \pm 4, y= \pm 3$) we draw the diagonals (these are the asymptotes of the hyperbola, i.e., the lines $y= \pm \frac{b}{a} x$; in our case, $y= \pm \frac{3}{4} x$).

The branches of the hyperbola pass through the points $(4,0),(-4,0)$, approaching the asymptotes, creating the impression of almost parallel lines. The foci $F_{1}(-5,0), \quad F_{2}(5,0)$ are considered to lie inside the hyperbola.

2) We have $\varepsilon=\frac{5}{4}$. The desired point $M(x, y)$ will be determined using the formula $r_{1}=2 r_{2}$, or

$$
\left|4+\frac{5}{4} x\right|=2\left|4-\frac{5}{4} x\right| \Leftrightarrow\left[\begin{array}{l}
4+\frac{5}{4} x=2\left(4-\frac{5}{4} x\right) \\
4+\frac{5}{4} x=-2\left(4-\frac{5}{4} x\right)
\end{array}\right.
$$

We find $x=\frac{48}{5}$ and $x=\frac{16}{15}$.

Since $M(x, y)$ lies on the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$, the ordinates of the corresponding points will be found from this equation for the found values of $x: y= \pm \frac{3}{4} \sqrt{x^{2}-16}$, and if $x=\frac{48}{5}$, then $y= \pm \frac{3}{4} \sqrt{\left(\frac{48}{5}\right)^{2}-16}= \pm \frac{3}{5} \sqrt{119}, \quad$ and if $\quad x=\frac{16}{15}, \quad$ then $y= \pm \frac{3}{4} \sqrt{\left(\frac{16}{15}\right)^{2}-16}= \pm \frac{3}{4} \sqrt{-\frac{3344}{225}}$ (this number does not exist in the required sense).

We obtained two points that satisfy the given conditions,

$$
M_{1}\left(\frac{48}{5}, \frac{3}{5} \sqrt{119}\right), \quad M_{2}=\left(\frac{48}{5},-\frac{3}{5} \sqrt{119}\right)
$$

3) The equations of the directrices of the given hyperbola: $x= \pm \frac{4}{5 / 4}= \pm \frac{16}{5}$.