Example 8. Determine the coordinates of the point of intersection of two mutually perpendicular lines passing through the foci of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$, given that the point $A(6,-2)$ lies on the line passing through its right focus.
Solution. 1) Let's make a drawing (Fig. 2.24) and list the parameters of the hyperbola. We have \(a=4, b=3, c=5, F_{1}(-5,0), F_{2}(5,0)\). We proceed to the calculations.

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-030.jpg?height=343&width=535&top_left_y=233&top_left_x=98)

Fig. 2.24

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-030.jpg?height=409&width=360&top_left_y=174&top_left_x=741)

Fig. 2.25
2) We will form the equation of \(A F_{2}\) using two points:

$$
\frac{x-5}{6-5}=\frac{y-0}{-2-0} \Rightarrow y=-2(x-5), \text { or } 2 x+y-10=0
$$

3) We will form the equation of the line \(F_{1} M\), passing through \(F_{1}\) and perpendicular to the line \(A F_{2}\). We have \(k_{A F_{2}}=-2\), so \(k_{F_{1} M}=1 / 2\). We get \((F_{1} M): y-0=1 / 2(x+5)\), or \(x-2 y+5=0\).
4) The coordinates of point \(M\) are obtained by solving the system

$$
\left\{\begin{array}{l}
2 x+y-10=0, \\
x-2 y+5=0
\end{array} \Rightarrow M(3,4)\right.
$$

## 4.4. Parabola

A parabola is the set of all points (GMT) for which the distance to a fixed point, called the focus, is equal to the distance to a fixed line, called the directrix. If the focus of the parabola is located at \(F\left(\frac{p}{2}, 0\right)\) and the directrix has the equation \(x=-\frac{p}{2}\), then such a parabola has the canonical equation \(y^{2}=2 p x\). Here, \(p\) is called the parameter of the parabola. The distance from a point \(M(x, y)\) on the parabola to the focus \(F\) is \(r=x+\frac{p}{2}\) (Fig. 2.25).

## Examples with Solutions