Polar Conics Example 2 - Sketch r = 12/(6+2sinθ)

In this video, we'll identify the features that will allow us to sketch the equation r = 12/(2+6sinθ).

To sketch the equation r = 12/(6+2sinθ), we first need to convert it to the standard polar form of:

r = \frac{l}{1+e\sin\theta}

We can do this by dividing both the top and bottom of this fraction by 6, leaving us with...

r = \frac{2}{1+\frac{1}{3}\sin\theta}

Now from this, we can see that this conic has an eccentricity e = ⅓,  which falls between 0 and 1, meaning it forms an ellipse.

From there, we can find the vertices by letting θ = π/2, which gives r(π/2) = 3/2; and θ = 3π/2, which gives r(3π/2) = 3.

The semi-latus rectum length l = 2, which means at θ = 0 and θ = π, directly to the right and left of the focus, the ellipse has r = 2.

To find the centre-point C and minor vertices, we need to find the semi-major axis length. This is simply the distance between the 2 major vertices divided by 2. The result is a = 9/4.

Then the distance from the centre-point C and the focus F is c = ae = (9/4)(1/3) = ¾.

Finally the minor vertices B and B' are at a semi-minor axis length of b either side of the centre-point. We find b from the equation:

b^2 = a^2(1 - e^2)

Thus b = 3/√2.