How will you find out if the point $\displaystyle (\alpha\beta)$ is outside, on or inside the ellipse $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$

Hence, show that the triangle whose vertices are $\displaystyle (1,2),(3,-1)$ and $\displaystyle (-2,1)$ lies wholly inside the ellipse $\displaystyle x^2+2y^2=13.$