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

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(C) The given equation is $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos 	heta$. Multiplying by $8r$,we get: $8 = r + 3r \cos 	heta$ Since $x = r \co

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a hyperbola

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(C) The given equation is $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos 	heta$. Multiplying by $8r$,we get: $8 = r + 3r \cos 	heta$ Since $x = r \cos 	heta$,we have $r = 8 - 3x$. Squaring both sides: $r^2 = (8 - 3x)^2$ $x^2 + y^2 = 64 + 9x^2 - 48x$ $8x^2 - y^2 - 48x + 64 = 0$. Comparing this with the general equation of a conic section $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$,here $A = 8$ and $C = -1$. Since $A$ and $C$ have opposite signs $(AC < 0)$,the equation represents a hyperbola.

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

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