Answer:
the answer is C
Explanation:
i did the assignment on edg
The answer is (c) ⇒ e = 3 and k = 1/2 , conic: hyperbola
∵ r = ek/(1 ± ecosФ)
∵ r = 3/(2 - 6cosФ) ⇒ ÷ 2 up and down
∴ r = (3/2)/(1 - 3cosФ)
∴ e = 3
∵ ek = 3/2 ⇒ 3k = 3/2
∴ k = 1/2
∵ e > 1
∴ The conic is hyperbola
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