R = 2 / (1 + sin θ)

Using the following relations:

R = sqrt (x^2 + y^2)
sin θ = y/R

R = 2 / (1 + y/R)
R(1 + y/R) = 2
R + y = 2
R = 2 - y
sqrt(x^2 + y^2) = 2 - y

Squaring both sides:
x^2 + y^2 = (2 - y)^2
x^2 + y^2 = 4 - 4y + y^2

x^2 + 4y - 4

8 0

3 years ago

Use the quadratic formula to find both solutions to the quadratic equation given below 4x^2+3x-1=0

Aneli [31]

4x² + 3x - 1 = 0
a=4 b=3 c=-1

Quadratic formula:
x = $\frac{-b +/- \sqrt{b^{2} -4ac} }{2a} = \frac{-3 +/- \sqrt{3^{2} - 4(4)(-1)}}{2(4)}$
= $\frac{-3 +/- \sqrt{9 + 16}}{8} = \frac{-3 +/- \sqrt{25}}{8} = \frac{-3 +/- 5}{8}$
= $\frac{-3 + 5}{8}$ , $\frac{-3 - 5}{8}$
= $\frac{2}{8}$ , $\frac{-8}{8}$
= $\frac{1}{4}$ , -1

7 0

3 years ago

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