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 $\pi$.

Answer 2

The change in variables s=x+3y, t=y to find the area of the ellipse x²+6xy+10y²≤1 yields; s² +t²≤ 1.

Substitution of variables

It follows from the equations given that;

• s = x +3y, x = s -3y
• t = y

Hence, it follows from substitution that;

• (s-3t)² +6(s-3t)t +10t² ≤ 1

• s² -6st +9t² +6st-18t² +10t² ≤ 1

• s²+t² ≤ 1

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