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3 years ago

(Q2) Determine the eccentricity, the type of conic, and the directrix for r = 9/1+cos theta

Mathematics

1 answer:

g100num [7]3 years ago

3 0

Answer:

The answer is (d) ⇒ eccentricity = 6 , hyperbola , directrix x = 3/2

Step-by-step explanation:

∵ r = ek/(1 ± ecosФ)

∵ r = 9/(1 + 6cosФ)

∴ e = 6 ⇒ eccentricity = 6

∵ e > 1

∴ The conic is hyperbola

∵ ek = 9

∴ k = 9/6 = 3/2

∴ The directrix x = 3/2

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Diagonal of the parallelogram divides the parallelogram in to two equal areas.

So area of parallelogram = 2(area of triangle)

According to the given diagram,

AB= 8, AD = 5 and BD = 11

So according to the Heron's formula,

Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}\\\\where,\\ s =\frac{a + b + c }{2}$

and a, b and c are the three sides of the triangle

Area of triangle ABD = $s=\frac{8 + 5 + 11}{2} \\\\s = \frac{24}{2}\\\\s = 12Area of triangle ABD = \sqrt{12(12 - 8) (12 - 5) (12 - 11)}\\\\Area of triangle ABD = \sqrt{12(4) (7) (1)}\\\\Area of triangle ABD = \sqrt{336}\\\\Area of triangle ABD = 18.33$

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2 years ago

Read 2 more answers