3. The line $y=ax+b$ passes through the point $T(0,6)$, with the coordinate axes it forms a triangle with an area of 24 and does not pass through the third quadrant.

a) Determine $a$ and $b$.

b) Determine the volume of the body obtained by rotating the resulting triangle around its longest side.
Solution. a) The triangle OST is a right triangle (drawing on the right) and its area is equal to 24, so

$$
24=P=\frac{1}{2} O T \cdot O S=3 O S
$$

from which it follows that $O S=8$. Therefore, the line $y=a x+b$ passes through the points $T(0,6)$ and

![](https://cdn.mathpix.com/cropped/2024_06_05_e381a8d26dc145f66c35g-4.jpg?height=337&width=414&top_left_y=776&top_left_x=1069)
$S(8,0)$, so $b=6$ and $8 a+b=0$, from which we find $a=-\frac{3}{4}$.

b) By rotating around the longest side, the triangle forms a double cone with radius $O H$ and height ST. We have,

$$
\begin{aligned}
& S T=\sqrt{O S^{2}+O T^{2}}=\sqrt{8^{2}+6^{2}}=10 \text { and } \\
& 24=P=\frac{S T \cdot O H}{2}=5 O H, \text { i.e., } O H=4.8
\end{aligned}
$$

For the surface area of the double cone, we have

$$
P=\pi \cdot O H \cdot O T+\pi \cdot O H \cdot O S=67.2 \pi
$$

and for the volume, we have

$$
V=\frac{\pi}{3} O H^{2} \cdot S H+\frac{\pi}{3} O H^{2} \cdot H T=\frac{\pi}{3} O H^{2} \cdot(S H+H T)=\frac{\pi}{3} O H^{2} \cdot S T=76.8 \pi
$$