identify the graph of the equation.
asked Sep 13, 2011 in Trigonometry Answers by anonymous

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This looks like the equation of an ellipse which has the standard form x²/a²+y²/b²=1, centrally located because there are no x or y terms in the given equation.

The xy term suggests a tilted axis. If the angle of tilt is ɸ, we can express x and y in polar coordinates:

x=rcosθ, y=rsinθ.

After tilting we get

x'=rcos(θ+ɸ)=rcosθcosɸ-rsinθsinɸ=xcosɸ-ysinɸ

y'=rsin(θ+ɸ)=rsinθcosɸ+rcosθsinɸ=ycosɸ+xsinɸ

Now we redo the standard equation using the substitutions x=x' and y=y':

b²(x²₋cos²ɸ-xysin(2ɸ)+y²sin²ɸ)+a²(y²₋cos²ɸ+xysin(2ɸ)+x²sin²ɸ)=a²b².

Now we equate coefficients for matching terms in the given equation:

a²b²=3, so b²=3/a²

xy term: (a²-b²)sin(2ɸ)=5, ɸ=½arcsin(5/(a²-b²))

x² term: b²cos²ɸ+a²sin²ɸ=9

y² term: b²sin²ɸ+a²cos²ɸ=1

Add these two equations: b²+a²=10, 3/a²+a²=10, a⁴-10a²=-3, a⁴-10a²+25=-3+25=22, (a²-5)²=22. So a²=5±√22.

From this b²=5∓√22, a²-b²=±2√22. Therefore ɸ=½arcsin(±5/(2√22))=±16.10425°, which is ±16° to the nearest degree.

 

answered Jun 21 by Rod Top Rated User (552,540 points)

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