A2 A point with coordinates $(a, 2 a)$ lies in the 3rd quadrant and on the curve given by the equation $3 x^{2}+y^{2}=28$. Find $a$.
Answer: -2
Solution:
We plug in the point $(x, y)=(a, 2 a)$ into the equation of the curve, to get
$$
3 a^{2}+(2 a)^{2}=28,
$$
and now expanding and dividing both sides of the equation by 7 , we get that $a^{2}=4$, and so $a= \pm 2$. Since the point lies in the third quadrant, we must choose $a=-2$.