Task B-4.5. If we draw a tangent $t_{1}$ to the ellipse $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$ at an angle of $45^{\circ}$ to the positive direction of the $x$-axis, its intercept on the $y$-axis is 4. If we draw a tangent $t_{2}$ at an angle of $60^{\circ}$, the intercept on the $y$-axis will increase by 2. Determine the area of the quadrilateral $F_{1} T_{1} F_{2} T_{2}$, where $F_{1}, F_{2}$ are the foci of the ellipse, and $T_{1}, T_{2}$ are the intersections of the tangents $t_{1}, t_{2}$ with the $y$-axis.
Solution. Let's write the equations of the given tangents:

$$
\begin{gathered}
t_{1} \ldots k_{1}=\operatorname{tg} 45^{\circ}=1, l_{2}=4, \quad t_{2} \ldots k_{2}=\sqrt{3}, l_{2}=6 \\
t_{1} \ldots y=x+4 \\
t_{2} \ldots y=\sqrt{3} x+6
\end{gathered}
$$

From the condition of tangency $a^{2} k^{2}+b^{2}=l^{2}$, it follows that $a^{2}+b^{2}=16$ and $3 a^{2}+b^{2}=36$, hence $a^{2}=10$, $b^{2}=6$.

(1 point)

Then $e^{2}=10-6=4$, and the coordinates of the foci are $F_{1}=(2,0), F_{2}=(-2,0)$.

Since $T_{1}=(0,4), T_{2}=(0,6)$, the desired area is

$$
P=2 \cdot\left(\frac{2 \cdot 6}{2}-\frac{2 \cdot 4}{2}\right)=4
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_0148fcdd1c92efebb536g-23.jpg?height=929&width=1070&top_left_y=1319&top_left_x=495)