Ellipse with axes tilted by 45 degrees

Text

An ellipse has its centre at the origin and axes tilted by $45^\circ$; its equation is $13X^2 - 10XY + 13Y^2 = 144.$

• (a) find the semi-axes;
• (b) write the ellipse with the same semi-axes but axes aligned with the coordinate axes;
• (c) verify that a $45^\circ$ rotation returns the given equation.

Solution

(a) Rotate by $45^\circ$ with $X=\dfrac{u+v}{\sqrt2},\ Y=\dfrac{u-v}{\sqrt2}$. Then $13(X^2+Y^2)=13(u^2+v^2)$ and $-10XY=-5(u^2-v^2)$, so $8u^2+18v^2=144 \implies \frac{u^2}{18}+\frac{v^2}{8}=1.$ The semi-axes are $a=\sqrt{18}=3\sqrt2\approx4{,}24$ (along the bisector $y=x$) and $b=\sqrt{8}=2\sqrt2\approx2{,}83$.

(b) Axis-aligned: $\dfrac{x^2}{18}+\dfrac{y^2}{8}=1$, i.e. $4x^2+9y^2=72$.

(c) Starting from $4u^2+9v^2=72$ and substituting $u=\dfrac{x+y}{\sqrt2},\ v=\dfrac{x-y}{\sqrt2}$ gives $4(x+y)^2+9(x-y)^2=144$, i.e. $13x^2-10xy+13y^2=144$. ✓

$\boxed{a=3\sqrt2,\qquad b=2\sqrt2}$