Rotating the conic $25x^2+25y^2-14xy-32\sqrt{2}x+32\sqrt{2}y-256=0$

Question

The answer I got is $α$ = $-π/4$.

But I'm not sure that this is this the correct answer because when I sub. in $α$ = $-π/4$ into

[25−14 cos(α)sin(α)]x˜^2, the coefficient doesn't turn out to be 16 but 32.

Answer 1

For $4\alpha=-\pi,$

$$\cos\alpha+\sin\alpha=0$$

$$\cos\alpha-\sin\alpha=\dfrac2{\sqrt2}$$

$$\cos\alpha\sin\alpha=-\dfrac12$$

We get $$(25+7)x^2+(25-7)y^2-64x-256=0$$

Divide both sides by $2$

Answer 2

You want $$\cos^2 \alpha-\sin^2 \alpha =0,$$ which implies $$\cos\alpha-\sin \alpha =0$$ or $$\cos\alpha+\sin\alpha =0.$$ The first of these possibilities already gives you what you want (namely that $\alpha=π/4$) since $0<π/4<π/2.$