Question

Use the change of variables s=x+3y, t=y to find the area of the ellipse x2+6xy+10y2≤1.

Answer 1

$t=y\implies s=x+3t\implies x=s-3t$

Then

$x^2+6xy+10y^2=(s-3t)^2+6(s-3t)t+10t^2=s^2+t^2\le1$

which is a disk of radius 1, hence with area $\pi$.