Trigonometric Conics

Geometry Level 4

If x 2 cos ⁡ ( θ ) − y 2 sin ⁡ ( θ ) = sin ⁡ ( 2 θ ) x^2\cos(\theta)-y^2\sin(\theta)=\sin(2\theta) , what is the distance from the graph's center to its focus?

Note: Take 180 ° < θ < 270 ° 180°<\theta<270° .


The answer is 2.

Yashas Ravi by Yashas Ravi , 18, USA

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1 solution

Yashas Ravi
Jun 24, 2019

The equation can be simplified to x 2 cos ⁡ ( θ ) − y 2 sin ⁡ ( θ ) = 2 sin ⁡ ( θ ) cos ⁡ ( θ ) x^2\cos(θ)-y^2\sin(θ)=2\sin(θ)\cos(θ) which becomes x 2 / 2 sin ⁡ ( θ ) − y 2 / 2 cos ⁡ ( θ ) = 1 x^2/2\sin(θ)-y^2/2\cos(θ)=1 . The distance from the center to the focus of this hyperbola is ( 2 sin ⁡ ( θ ) ) 2 + ( 2 cos ⁡ ( θ ) ) 2 = 4 ( sin ⁡ ( θ ) 2 + cos ⁡ ( θ ) 2 ) = 4 = 2 \sqrt{(2\sin(θ))^2+(2\cos(θ))^2} = \sqrt{4(\sin(θ)^2+\cos(θ)^2)} = \sqrt{4} = 2 which is the final answer.

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From what I can see, the solution is wrong.

Atomsky Jahid - 1 year, 11 months ago

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There's a mistake in the formula for the distance. In general, for the hyperbola x 2 a 2 − y 2 b 2 = 1 \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 the distance from the centre to the foci is a 2 + b 2 \sqrt{a^2+b^2} .

In this case, we have a 2 = 2 sin ⁡ θ a^2=2\sin \theta and b 2 = 2 cos ⁡ θ b^2=2\cos \theta (note that the right-hand sides of these should not be squared again, as in the given solution). This gives an answer of 2 ( sin ⁡ θ + cos ⁡ θ ) \boxed{\sqrt{2(\sin \theta+\cos \theta)}} , which varies with θ \theta .

Chris Lewis - 1 year, 9 months ago

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