How do you graph r=4−3+3sinθ?

1 Answer
Douglas K.
Oct 15, 2016

This is a hyperbola with its center (0,−4); its vertices are at (0,−4−√12)and(0,−4+√12) and it opens downward and upward from its vertices.

Explanation:

Multiply both sides by the denominator:

4rsin(θ)−3r=4

Multiply both side by -1:

3r−4rsin(θ)=−4

Move the sine term to the right:

3r=4rsin(θ)−4

Factor out a 4:

3r=4(rsin(θ)−1)

Square both sides:

9r2=16(rsin(θ)−1)2

Substitute x2+y2 for r2 and y for rsin(θ)

9(x2+y2)=16(y−1)2

Expand the square on the right:

9(x2+y2)=16(y2−2y+1)

distribute through the ()s:

9x2+9y2=16y2−32y+16

Combine like terms and leave the constant on the right:

9x2−4y2−32y=16

Add −4k2 to both sides:

9x2−4y2−32y−4k2=16−4k2

Factor out -4 from the y terms

9x2−4(y2+8y+k2)=16−4k2

Use 8y to find the value of k and k2:

−2ky=8y

k=−4,k2=16

This makes the left a perfect square with k:

9x2−4(y−−4)2=−48

Divide both sides by -48:

(y−−4)212−316x2=1

Put in standard form:

(y−−4)2(√12)2−(x−0)2(4√33)2=1

This is a hyperbola with its center (0,−4) its vertices are at (0,−4−√12)and(0,−4+√12). It opens downward and upward from its vertices.

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