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Math  /  Geometry

Question1. Convert the cartesian point (−23,2)(-2 \sqrt{3}, 2) into its polar form. List all possible representations.
2. Write the equation of the circle (x−3)2+(y+4)2=25(x-3)^{2}+(y+4)^{2}=25 into its polar form r=ρ(θ)r=\rho(\theta).
3. Write the polar equation of the following polar graph:

Studdy Solution

STEP 1

1. The point (−23,2)(-2 \sqrt{3}, 2) is given in Cartesian coordinates.
2. The polar form of a point is represented as (r,θ)(r, \theta).
3. The circle equation (x−3)2+(y+4)2=25(x-3)^2 + (y+4)^2 = 25 needs to be converted to polar coordinates.
4. Polar coordinates are related to Cartesian coordinates by the equations x=rcos⁡θx = r \cos \theta and y=rsin⁡θy = r \sin \theta.
5. The polar equation of a circle can be expressed in the form r=ρ(θ)r = \rho(\theta).

STEP 2

1. Convert the Cartesian point to polar form.
2. Convert the Cartesian equation of the circle to polar form.
3. Write the polar equation of the given polar graph.

STEP 3

To convert the Cartesian point (−23,2)(-2 \sqrt{3}, 2) to polar form, calculate the radius rr using the formula:
r=x2+y2 r = \sqrt{x^2 + y^2}
Substitute x=−23x = -2 \sqrt{3} and y=2y = 2:
r=(−23)2+22 r = \sqrt{(-2 \sqrt{3})^2 + 2^2} =12+4 = \sqrt{12 + 4} =16 = \sqrt{16} =4 = 4

STEP 4

Calculate the angle θ\theta using the formula:
θ=tan⁡−1(yx) \theta = \tan^{-1}\left(\frac{y}{x}\right)
Substitute x=−23x = -2 \sqrt{3} and y=2y = 2:
θ=tan⁡−1(2−23) \theta = \tan^{-1}\left(\frac{2}{-2 \sqrt{3}}\right) =tan⁡−1(−13) = \tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) =tan⁡−1(−33) = \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)
The reference angle is π6\frac{\pi}{6}. Since the point is in the second quadrant, θ=π−π6=5π6\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}.

STEP 5

List all possible polar representations:
The point can be represented as (4,5π6)(4, \frac{5\pi}{6}) or (−4,11π6)(-4, \frac{11\pi}{6}).

STEP 6

Convert the circle equation (x−3)2+(y+4)2=25(x-3)^2 + (y+4)^2 = 25 to polar form.
Substitute x=rcos⁡θx = r \cos \theta and y=rsin⁡θy = r \sin \theta:
(rcos⁡θ−3)2+(rsin⁡θ+4)2=25(r \cos \theta - 3)^2 + (r \sin \theta + 4)^2 = 25
Expand and simplify:
(rcos⁡θ−3)2=r2cos⁡2θ−6rcos⁡θ+9 (r \cos \theta - 3)^2 = r^2 \cos^2 \theta - 6r \cos \theta + 9 (rsin⁡θ+4)2=r2sin⁡2θ+8rsin⁡θ+16 (r \sin \theta + 4)^2 = r^2 \sin^2 \theta + 8r \sin \theta + 16
Combine terms:
r2(cos⁡2θ+sin⁡2θ)−6rcos⁡θ+8rsin⁡θ+25=25 r^2 (\cos^2 \theta + \sin^2 \theta) - 6r \cos \theta + 8r \sin \theta + 25 = 25
Since cos⁡2θ+sin⁡2θ=1\cos^2 \theta + \sin^2 \theta = 1:
r2−6rcos⁡θ+8rsin⁡θ=0 r^2 - 6r \cos \theta + 8r \sin \theta = 0
Factor out rr:
r(r−6cos⁡θ+8sin⁡θ)=0 r (r - 6 \cos \theta + 8 \sin \theta) = 0
Since r=0r = 0 is trivial, the equation becomes:
r=6cos⁡θ−8sin⁡θ r = 6 \cos \theta - 8 \sin \theta

STEP 7

Write the polar equation of the given polar graph.
This step requires additional information about the specific polar graph to provide an accurate equation. Please provide details of the polar graph.

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