In this set of supplemental notes I will demonstrate how to convert equations of lines, circles, and conic sections into their related polar form. Some of the polar forms will be useful, while others a waste of time. Let us start out discussion with the topic - the polar equation for a line.

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Suppose we want to write a polar equation of a line, so let us derive out the formula for this equation.

figure 1

Suppose that a point P₀ (r ₀, q ₀), r ₀ ³ 0 is on the line, and let there be a perpendicular from the origin to the line L meet at this point. (see figure 1)

If P (r, q ) is any other point on L, the points P, P₀, and O are vertices of a right triangle. Therefore,

or r ₀ = r cos (q - q ₀).

r ₀ = r cos (q - q ₀)

EXAMPLE 1:

Find the polar and Cartesian equations for the line through the point (3, 4p /3).

SOLUTION:

For this example, r ₀ = 3 and q ₀ = 4p / 3, so plug these values into the formula.

3 = r cos (q - 4p / 3)

Now to convert the polar equation to it's equivalent Cartesian equation using the following trig identity.

FACT: cos (A - B) = cos A cos B + sin A sin B

EXAMPLE 2:

Find the polar and Cartesian equations for the line through the point

(2, - p / 6).

SOLUTION:

For this example, r ₀ = 2 and q ₀ = - p / 6 so plug these values into the formula.

2 = r cos (q - (-p / 6)) ® 2 = r cos (q + p / 6)

Now to convert the polar equation to it's equivalent Cartesian equation using the following trig identity.

FACT: cos (A + B) = cos A cos B - sin A sin B

EXAMPLE 3:

Find a polar equation in the form r cos (q - q ₀) = r ₀ for the line

SOLUTION:

First, let us convert the equation into its polar form.

From this conversion and simplification we have determined that r ₀ = 3. Now we need to determine what q ₀ is. To do this, we need to find the slope of this line.

The slope of this line is m = -1.

What is the slope of the perpendicular to this line?

If m ₁ = -1, then m ₂ = 1.

Remember that tangent of q ₀ is equal to the perpendicular slope.

tan q ₀ = 1 ® q ₀ = p / 4

Using all that data that we have found, we can now write the polar equation for this line. Here is the equation.

3 = r cos (q - p / 4)

EXAMPLE 4:

Find the polar equation of the form r cos (q - q ₀) = r ₀ for the line x = - 4.

SOLUTION:

x = - 4 is an equation of a vertical line, so just convert this equation into it's corresponding polar form.

x = - 4 ® r cos q = -4

Now let us derive the equation for a circle of radius a centered at P₀ (r ₀, q ₀).

figure 2

Let P (r, q ) be a point on the circle and apply the law of cosines to D OP ₀ P.

If the circle passes through the origin, then r ₀ = a and

If the circle's center lies on the positive x-axis, then q ₀ = 0 and r = 2a cos q .

If the circle's center lies on the positive y-axis, then q ₀ = p / 2 and cos (q - p / 2) = sin q and r = 2a sin q .

r = 2a cos q

r = - 2a cos q

r = 2a sin q

r = - 2a sin q

EXAMPLE 5:

Find the polar equation of the circle with center (2, p / 2).

Since r ₀ = a = 2 > 0, the equation that I will use is r = 2a sin q .

r = 2( 2) sin q ® r = 4 sin q

Here is the graph of this circle.

EXAMPLE 6:

Find the polar equation of the circle with center at (-1, 0).

SOLUTION:

Since r ₀ = a = -1 < 0, the equation that I will use is r = -2a cos q and I will use the absolute value of a.

r = -2(1) cos q ® r = -2 cos q

Here is the graph of this circle.

EXAMPLE 7:

Given the polar coordinates for the center and identify the radius of the circle r = 6sin q .

SOLUTION:

This circle is of the form r = 2a sin q , so 2a = 6 or a = 3. Since this equation contains sine, and not cosine, this implies that q ₀ = p / 2. Therefore the center is (3, p / 2) and the radius is 3.

Here is the graph of the circle. Does the answers fit the graph?

EXAMPLE 8:

Find the polar equation for the circle x ² + (y - 2) ² = 4.

SOLUTION:

First expand x ² + (y - 2) ² = 4.

x ² + (y - 2) ² = 4 ® x ² + y ² - 4y + 4 = 4 ® x ² + y ² - 4y = 0

Now, convert it to its polar form.

x ² + y ² - 4y = 0 ® r ² - 4 r sin q = 0 ® r ² = 4 r sin q

® r = 4sin q

Here is the graph of this circle.

EXAMPLE 9:

Find the polar equation for the circle x ² + y ² + x = 0.

SOLUTION:

Just go ahead and convert this equation into it's polar form.

x ² + y ² + x = 0 ® r ² + r cos q = 0 ® r ² = - r cos q

® r = - cos q

Here is the graph of this circle.

figure 3

To find the polar equations for ellipses, parabolas, and hyperbolas, we will place one of the foci at the origin and the corresponding directrix to the right of the origin along the vertical line x = k. (see figure 3)

Setting the problem up this way makes the line segment PF = r and the line segment PD = k - FB = k - r cos q .

Early in chapter 9, the authors of the text stated that the conic's focus-directrix equation is PF = (e)(PD), where e is the eccentricity.

Using this equation, substitute in the expressions for PF and Pd.

where e > 0 and x = k

This is the basic polar equation for a conic section. There are three more equations that are based on where the directrix lies.

Recall how we can use the eccentricity to determine the type of conic section we are working with. Here is a quick review of this topic.

If e = 0, then conic section is a circle (special case of an ellipse).

If 0 £ e < 1, the conic section is an ellipse

If e = 1, the conic section is a parabola.

If e > 1, the conic section is a hyperbola.

EXAMPLE 10:

Given the eccentricity, e = 1/4, and directrix x = -2, find the polar equation for this conic section.

SOLUTION:

Since x = -2, this tells me that I need to use the formula

For this problem, e = 1/4 and k = 2, so plug these into the formula.

Here is the graph of this ellipse.

EXAMPLE 11:

Given the eccentricity, e = 2 and directrix y = 4, find the polar equation for this conic section.

SOLUTION:

The directrix is of the form y = k, so I will use the formula

For this problem, e = 2 and k = 4, so plug these values into the formula.

Here is the graph of this hyperbola.