Example 7. On the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$, find the point $M(x, y)$ such that its distance to one asymptote is three times greater than the distance to the other asymptote.
Solution. 1) Let's make a symbolic drawing of the hyperbola (Fig. 2.22) and its asymptotes. It depicts two different possible situations that satisfy the conditions of the problem: the distance from point $M$ to asymptote $l_{2}$ is three times greater than the distance to asymptote $l_{1}$; for point $M_{\mathrm{I}}-$, it is the opposite.

2) Equations of the asymptotes:

$$
\begin{gathered}
l_{1}: y=\frac{3}{4} x, 3 x-4 y=0 \\
l_{2}: y=-\frac{3}{4} x, 3 x+4 y=0
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-029.jpg?height=339&width=473&top_left_y=438&top_left_x=2010)

Fig. 2.23
3) For point $M_{1}(x, y)$, we have $d\left(M_{1}, l_{1}\right)=3 d\left(M_{1}, l_{2}\right)$. According to the corresponding formulas ( $\S 3$, point $5^{\circ}$ ), this equality can be rewritten as

$$
\begin{gathered}
\frac{|3 x-4 y|}{\sqrt{3^{2}+(-4)^{2}}}=\frac{3|3 x+4 y|}{\sqrt{3^{2}+4^{2}}}, \quad|3 x-4 y|=3|3 x+4 y| \Leftrightarrow \\
\Leftrightarrow\left[\begin{array} { l } 
{ 3 x - 4 y = 9 x + 1 2 y , } \\
{ 3 x - 4 y = - 9 x - 1 2 y , }
\end{array} \quad \left[\begin{array}{l}
3 x=-8 y \\
3 x=-2 y
\end{array}\right.\right.
\end{gathered}
$$

4) Since $M_{1}$ lies on the hyperbola, we need to solve the following systems

$$
\left\{\begin{array} { l } 
{ \frac { x ^ { 2 } } { 1 6 } - \frac { y ^ { 2 } } { 9 } = 1 , } \\
{ 3 x = - 8 y }
\end{array} \quad \text { and } \quad \left\{\begin{array}{l}
\frac{x^{2}}{16}-\frac{y^{2}}{9}=1 \\
3 x=-2 y
\end{array}\right.\right.
$$

From the first system, we find $x= \pm \frac{8}{\sqrt{3}}, y=\mp \sqrt{3}$, which corresponds to two points $M_{1}\left(-\frac{8}{\sqrt{3}}, \sqrt{3}\right), M_{1}^{\prime}\left(\frac{8}{\sqrt{3}},-\sqrt{3}\right)$.

The second system has no solutions.

5) As for the coordinates of point $M$, we suggest verifying independently that $M\left(\frac{8}{\sqrt{3}}, \sqrt{3}\right), M^{\prime}\left(-\frac{8}{\sqrt{3}},-\sqrt{3}\right)$.