Question

The point $(3,4)$ lies

Answer 1

For the point (3, 4)
1)$\frac{x^2}{18}+\frac{y^2}{32}-1=\frac{9}{18}+\frac{16}{32}-1=0$
So the point (3, 4) lies on the curve
2) $x^2+y^2-6x-8y=0$ is a circle with centre $(3,4)$
3) $y^2-8x=16-24<0$
$\Rightarrow$ The point (3, 4) lies inside the parabola $y^2=8x$.
4) $x^2+y^2-16=9+16-16>0$
$\Rightarrow$ The points lies outside the circle $x^2+y^2=16$.