Question

Find the set of possible of $\alpha$ for which point $P(\alpha ,3\alpha )$ lies on the smaller region of the ellipse $9x^2+16y^2=144$ divided by the line $3x+4y=12$.

Answer 1

$\frac{x^2}{16}+\frac{y^2}{9}=1$ $3x+4y=12$

Smaller region

$3x+4y-12>0$ and $\frac{x^2}{16}+\frac{y^2}{9}-1<0$

$P(\alpha ,3\alpha )$ $\frac{{\alpha }^2}{16}+{\alpha }^2-1<0$

$3\alpha +12\alpha -12>0$ $\frac{17{\alpha }^2}{16}<1$

$5\alpha -4>0$

$\alpha >\frac{4}{5}$ ${\alpha }^2<\frac{16}{17}$

$\alpha \in (\frac{4}{5},\frac{4}{\sqrt{17}})$