The equation frac{1}{r} = frac{1}{8} + frac{3 | vedclass

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(C) Given the polar equation: $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos 	heta$ Multiply by $8r$: $8 = r + 3r \cos 	heta$ Since $x = r \cos 	het

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a hyperbola

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(C) Given the polar equation: $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos 	heta$ Multiply by $8r$: $8 = r + 3r \cos 	heta$ Since $x = r \cos 	heta$ and $r = \sqrt{x^2 + y^2}$,we have: $r = 8 - 3x$ Squaring both sides: $r^2 = (8 - 3x)^2$ $x^2 + y^2 = 64 - 48x + 9x^2$ Rearranging the terms: $8x^2 - y^2 - 48x + 64 = 0$ This is of the form $Ax^2 + By^2 + Dx + Ey + F = 0$ where the coefficients of $x^2$ and $y^2$ have opposite signs ($8$ and $-1$). Therefore,it represents a hyperbola.

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents

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