16. The diagram shows the ellipse whose equation is $x^{2}+y^{2}-x y+x-4 y=12$. The curve cuts the $y$-axis at points $A$ and $C$ and cuts the $x$-axis at points $B$ and $D$.
What is the area of the inscribed quadrilateral $A B C D$ ?
A 28
B 36
C 42
D 48
E 56
Solution: A
The graph cuts the $y$-axis at points where $x=0$, and hence where $y^{2}-4 y=12$, that is, where $y^{2}-4 y-12=0$, that is, $(y+2)(y-6)=0$, giving $y=-2$ and $y=6$. Hence $A$ is the point $(0,6)$, $C$ is the point $(0,-2)$ and so $A C$ has length 8 . Similarly, $B$ and $D$ are points where $y=0$ and hence $x^{2}+x-12=0$, that is, $(x+4)(x-3)=0$, giving $x=-4$ and $x=3$. So $B$ is the point $(3,0), D$ is the point $(-4,0$ ), and so $B D$ has length 7. The area of the inscribed quadrilateral $A B C D$ is therefore $\frac{1}{2} A C . B D=\frac{1}{2}(8 \times 7)=28$.