Multiply both sides by the denominator:
4rsin(theta) - 3r = 44rsin(θ)−3r=4
Multiply both side by -1:
3r - 4rsin(theta) = -43r−4rsin(θ)=−4
Move the sine term to the right:
3r = 4rsin(theta)-43r=4rsin(θ)−4
Factor out a 4:
3r = 4(rsin(theta)-1)3r=4(rsin(θ)−1)
Square both sides:
9r^2 = 16(rsin(theta)-1)^29r2=16(rsin(θ)−1)2
Substitute x^2 + y^2x2+y2 for r^2r2 and yy for rsin(theta)rsin(θ)
9(x^2 + y^2) = 16(y - 1)^29(x2+y2)=16(y−1)2
Expand the square on the right:
9(x^2 + y^2) = 16(y^2 - 2y + 1)9(x2+y2)=16(y2−2y+1)
distribute through the ()s:
9x^2 + 9y^2 = 16y^2 - 32y + 169x2+9y2=16y2−32y+16
Combine like terms and leave the constant on the right:
9x^2 - 4y^2 - 32y = 169x2−4y2−32y=16
Add -4k^2−4k2 to both sides:
9x^2 - 4y^2 - 32y -4k^2 = 16 - 4k^29x2−4y2−32y−4k2=16−4k2
Factor out -4 from the y terms
9x^2 - 4(y^2 + 8y + k^2) = 16 - 4k^29x2−4(y2+8y+k2)=16−4k2
Use 8y to find the value of kk and k^2k2:
-2ky = 8y−2ky=8y
k = -4, k^2 = 16k=−4,k2=16
This makes the left a perfect square with k:
9x^2 - 4(y - -4)^2 = -489x2−4(y−−4)2=−48
Divide both sides by -48:
(y - -4)^2/12 - 3/16x^2 = 1(y−−4)212−316x2=1
Put in standard form:
(y - -4)^2/(sqrt(12))^2 - (x - 0)^2/(4sqrt(3)/3)^2 = 1(y−−4)2(√12)2−(x−0)2(4√33)2=1
This is a hyperbola with its center (0, -4)(0,−4) its vertices are at (0, -4 - sqrt(12)) and (0, -4 + sqrt(12))(0,−4−√12)and(0,−4+√12). It opens downward and upward from its vertices.
