Example 2. Derive the equation of an ellipse that is symmetric with respect to the coordinate axes and passes through the points $A\left(3,-\frac{16}{5}\right)$ and $B\left(-4,-\frac{12}{5}\right)$. Find the distances from point $A$ to the foci. Find the eccentricity of the ellipse. Derive the equations of its directrices. Draw the diagram.
Solution. 1) The parameters $a$ and $b$ of the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ will be found by substituting the coordinates of points $A$ and $B$ into this equation. This leads to the system

$$
\left\{\begin{array} { l } 
{ \frac { 3 ^ { 2 } } { a ^ { 2 } } + \frac { ( - \frac { 1 6 } { 5 } ) ^ { 2 } } { b ^ { 2 } } = 1 , } \\
{ \frac { 4 ^ { 2 } } { a ^ { 2 } } + \frac { ( - \frac { 1 2 } { 5 } ) ^ { 2 } } { b ^ { 2 } } = 1 , }
\end{array} \quad \text { or } \quad \left\{\begin{array}{l}
\frac{9}{a^{2}}+\frac{256}{25 b^{2}}=1 \\
\frac{16}{a^{2}}+\frac{144}{25 b^{2}}=1
\end{array}\right.\right.
$$

After multiplying the first equation by 16 and the second by -9 and adding the results, we have

$$
\frac{256 \cdot 16}{25 b^{2}}-\frac{144 \cdot 9}{25 b^{2}}=7, \quad \text { or } \quad 175 b^{2}=2800
$$

From this, considering $b>0$, we find $b=4$, and then $a=5$.

The canonical equation of the ellipse is found: $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$.

2) The focal distance $c=\sqrt{a^{2}-b^{2}}=\sqrt{25-16}=3$.
3) The eccentricity is $\varepsilon=$ $=\frac{3}{5}=0.6$.
4) The distances from $A$ to the foci: $r_{1}=5+0.6 \cdot 3=6.8$; $r_{2}=5-0.6 \cdot 3=3.2$.
5) The equations of the directrices: $x=$ $= \pm \frac{5}{0.6}$, i.e., $x=-\frac{25}{3}$ (left), $x=\frac{25}{3}$ (right).

![](https://cdn.mathpix.com/cropped/2024_05_22_8f5be56c3f01f571fd60g-027.jpg?height=320&width=554&top_left_y=1534&top_left_x=1904)

Fig. 2.19

The drawing is constructed (Fig. 2.19).