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.
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Note that the given equation is symmetric in $x$ and $y$ – J. W. Tanner Aug 18 '19 at 20:09
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@J.W.Tanner Only almost symmetric -- the linear terms are actually antisymmetric. – Allawonder Aug 18 '19 at 20:54
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Examine all of the coefficients after the rotation. You might notice that they have a common factor. – amd Aug 20 '19 at 01:40
2 Answers
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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 $2$
lab bhattacharjee
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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 $0<π/4<π/2.$
Allawonder
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@CubbyKushi Your solution $-π/4$ also does nicely, since it satisfies the second of the possibilities above. This is not surprising geometrically, since rotating the axes of coordinates clockwise by $\phi$ always corresponds to rotating the plane anticlockwise by the same amount, and vice versa. – Allawonder Aug 18 '19 at 20:56