8. Find the equations of the tangents drawn from the point $M(0, -2)$ to the parabola $8 y=(x-3)^{2}$. Determine the angle between the tangents. Find the area of the triangle $A B M$, where $A$ and $B$ are the points of tangency.
# Solution:

$8 y=(x-3)^{2} ; M(0 ;-2)$.

$y=\frac{\left(x_{0}-3\right)^{2}}{8}+\frac{\left(x_{0}-3\right)}{4}\left(x-x_{0}\right)$

$y=\frac{9}{2}+\frac{\left(x_{0}-3\right)}{4} x-\frac{x_{0}^{2}}{8}$

$-2=\frac{9}{8}+\frac{\left(x_{0}-3\right)}{4} \cdot 0-\frac{x_{0}^{2}}{8}, x_{0}^{2}-25=0$,

$\left(x_{0}\right)_{1}=-5,\left(y_{0}\right)_{1}=8, A(-5 ; 8)$.

$\left(x_{0}\right)_{2}=5,\left(y_{0}\right)_{2}=0.5, B(5 ; 0.5)$.

1) $y=-2 x-2$; 2) $y=\frac{1}{2} x-2$.

The angle between the tangents is $90^{\circ}$.

$$
\begin{aligned}
& |A M|=\sqrt{(0+5)^{2}+(-2-8)^{2}}=\frac{5 \sqrt{5}}{2} ; \\
& |B M|=\sqrt{(0-5)^{2}+\left(-2-\frac{1}{2}\right)^{2}}=\frac{5 \sqrt{5}}{2} \cdot S_{\triangle A M B}=\frac{1}{2} \cdot A M \cdot B M=\frac{1}{2} \cdot 5 \sqrt{5} \cdot \frac{5 \sqrt{5}}{2}=\frac{125}{4} .
\end{aligned}
$$

Verification by areas: $S_{\triangle A M B}=10 \cdot 10-\frac{1}{2} \cdot 10 \cdot 7.5-\frac{1}{2} \cdot 5 \cdot 2.5-\frac{1}{2} \cdot 5 \cdot 10=\frac{125}{4}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_797a639488ab43e0ad60g-20.jpg?height=778&width=797&top_left_y=1210&top_left_x=1014)

Answer: $y=-2 x-2, y=\frac{1}{2} x-2 ; 90^{\circ} ; \frac{125}{4}$.