Sketch a graph of r=12/2+sin(θ). Use the graph to write a Cartesian equation for the conic. Find the standard form equation for a hyperbola with vertices at (6,0)and (−6,0) and asymptote y=5/3x

Question

Sketch a graph of . Use the graph to write a Cartesian equation for the conic.

Find the standard form equation for a hyperbola with vertices at 

and and asymptote 

Write a polar equation for a conic having focus at the origin, directrix , and eccentricity .

.

Answer 1

For r = 12/(2 + sin θ), as θ = 0º, 90º, 180º, and 270º, we get points (6, 0), (0, 4), (-6, 0), (0, -12).

The curve is convex. Plotting for all θ, we get what appears to be an ellipse centered at (0, -4). The y-difference is 16, and 1/2 8 16 = 8

Thus, the form of the ellipse will x^2/a^2 + (y+4)^2/64 = 1

Substituting in (6, 0), we get 6^2/a^2 + (4)^2/64 = 1

36/a^2 + 1/4 = 1

36/a^2 = 3/4

144 = 3a^2

a^2 = 48

x^2/48 + (y+4)^2/64 = 1

We can verify that this is correct.

If we cross-multiply the original equation, we get that 2r + r sin θ= 12, or 2r + y = 12, which may be rewritten

2√(x² + y²) + y = 12, or

(12 - y) = 2√(x² + y²)

Squaring both sides, we get 144 - 24y + y² = 4x² + 4y²

144 = 4x² + 4y² + 24y - y²

144 = 4x² + 4y² + 24y - y²

144 = 4x² + 3y² + 24y

Completing the square,

144 = 4x² + 3y² + 24y + 48 - 48

192 = 4x² + 3(y + 4)^2

1 = x²/ 48 + (y + 4)^2/64

This is just what we expected.

If the hyperbola has (6, 0) and (-6, 0), it is of the form

x^2/6^2 - y^2/b^2 = 1

As an asymptote is y = 5/3x, b/6 = 5/3

b = 10

x^2/6^2 - y^2/10^2 = 1

x^2/36 - y^2/100 = 1

Eccentricity = 2, so e = 2

As the origin and directrix are 2 apart, c = 2

e = c/a

2 = 2/a

a = 1

c = √(a^2+b^2)

c^2 = a^2+b^2

b^2 = c^2 - a^2 = 2^2 - 1^2 = 3

Since the focus is at the origin, we subtract 0 from y in the formula. Since the directrix is at x = 2, we subtract 2 from x in the formula

Thus,

(x-2)^2/1 - y^2/3 = 1