8. Find the equations of the tangents drawn from the point $M(3 ; 0)$ to the parabola $8 y=x^{2}+16$. Determine the angle between the tangents. Find the area of triangle $A B M$, where $A$ and $B$ are the points of tangency.
# Solution:

$8 y=x^{2}+16 ; M(3 ; 0)$.

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

$y=2+\frac{x_{0}}{4} x-\frac{x_{0}^{2}}{8}$.

$0=2+\frac{x_{0}}{4} \cdot 3-\frac{x_{0}^{2}}{8}, x_{0}^{2}-6 x_{0}-16=0$,

$\left(x_{0}\right)_{1}=-2,\left(y_{0}\right)_{1}=2.5, A(-2 ; 2.5)$.

$\left(x_{0}\right)_{2}=8,\left(y_{0}\right)_{2}=10, B(8 ; 10)$.

1) $y=-\frac{1}{2} x+\frac{3}{2}$; 2) $y=2 x-6$.

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

![](https://cdn.mathpix.com/cropped/2024_05_06_797a639488ab43e0ad60g-26.jpg?height=745&width=805&top_left_y=1281&top_left_x=969)

$|A M|=\sqrt{(3+2)^{2}+\left(0-\frac{5}{2}\right)^{2}}=\frac{5 \sqrt{5}}{2} ;|B M|=\sqrt{(3-8)^{2}+(0-10)^{2}}=5 \sqrt{5}$.

$S_{\triangle A M B}=\frac{1}{2} \cdot A M \cdot B M=\frac{1}{2} \cdot \frac{5 \sqrt{5}}{2} \cdot 5 \sqrt{5}=\frac{125}{4}$.

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

Answer: $y=-\frac{1}{2} x+\frac{3}{2}, y=2 x-6 ; 90^{\circ} ; \frac{125}{4}$.