Task A-4.7. (10 points)

Among all points $z$ in the complex plane for which $|z+3|+|z-3|=10$, determine the one that is closest to the line passing through the points $(-3+4i)$ and $(-8+i)$.
## Second Solution.

As in the first solution, we determine the equation of the ellipse $16 x^{2}+25 y^{2}=400$,

and the slope of the given line $k=\frac{3}{5}$.

We will find the desired point as the point of tangency of one of the tangents to the given ellipse that is parallel to the given line.

To determine the equations of the tangents with slope $k=\frac{3}{5}$, we use the condition of tangency $a^{2} k^{2}+b^{2}=l^{2}$.

$$
l^{2}=25 \cdot\left(\frac{3}{5}\right)^{2}+16=9+16=25
$$

Thus, the lines $y=\frac{3}{5} x+5$ and $y=\frac{3}{5} x-5$ are tangents.

From the diagram, we see that we need the "upper" tangent, $y=\frac{3}{5} x+5$.

(1 point)

Let's determine its point of tangency with the ellipse. Solving the system:

$$
\left\{\begin{array}{c}
16 x^{2}+25 y^{2}=400 \\
y=\frac{3}{5} x+5
\end{array}\right.
$$

We get the equation: $16 x^{2}+(3 x+25)^{2}=400$

which simplifies to $x^{2}+6 x+9=0$,

so the solution to the system is $x=-3, y=\frac{16}{5}$.

The desired point is $\left(-3, \frac{16}{5}\right)$.

Note. We can also derive the equation of the ellipse as follows:

$$
\begin{gathered}
|(x+i y)+3|+|(x+i y)-3|=10 \\
\sqrt{(x+3)^{2}+y^{2}}+\sqrt{(x-3)^{2}+y^{2}}=10 \\
\sqrt{(x+3)^{2}+y^{2}}=10-\sqrt{(x-3)^{2}+y^{2}} \\
(x+3)^{2}+y^{2}=100-20 \sqrt{(x-3)^{2}+y^{2}}+(x-3)^{2}+y^{2} \\
10 \sqrt{(x-3)^{2}+y^{2}}=50-6 x \\
100\left(x^{2}-6 x+9\right)=2500-600 x+36 x^{2} \\
16 x^{2}+25 y^{2}=400
\end{gathered}
$$