Convert from polar form to rectangular form.
(1) 2 sin(theta)- 3 cos(theta) = r
(2) rsin(theta+ pi/4) = 6
(1) 2 sin(theta)- 3 cos(theta) = r
(2) rsin(theta+ pi/4) = 6
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Use the standard eqations: x = r*cos(T); y = r*cos(T); r^2 = x^2 + y^2; tan(T) = y/x.
(1) sin(T) is, then, y/r; analogously, cos(T) = x/r.
So, 2y/r - 3x/r = r; multiply each side by r.
2y - 3x = r^2 = x^2 + y^2.
Now, we have 2y - 3x = x^2 + y^2. We could be done, but we probably want to make this look like something, so let's complete the square to get it into the standard conic equation form.
x^2 + 3x + y^2 - 2y = 0.
So, to complete the square for x, we add (b/2)^2 to each side; same for y, so:
x^2 + 3x + 9/4 + y^2 - 2y + 1 = 9/4 + 1.
Now, factor according to how we complete the square:
(x + 3/2)^2 + (y - 1)^2 = 13/4.
What you should see, now, is that, as the "x^2" and "y^2" terms are both multiplied by the same number (i.e. 1), then we have a circle with center at (-3/2, 1) and radius sqrt(13)/2. Remember that the standard form of a circle is (x - h)^2 + (y - h)^2 = R^2, where (h, k) is the center and R the radius.
(2) sin(T + pi/4) can be expanded by the trig. addition formula sin(A + B) = sinAcosB - cosAsinB. So, we simply get sqrt(2)/2(sinT + cos T). Now, if we multiply by r, we get sqrt(2)/2(y + x) = 6. We may manipulate this equation further, but you should see, by now, that we have a line. The slope-intercept form is y = -x + 6sqrt(2).
(1) sin(T) is, then, y/r; analogously, cos(T) = x/r.
So, 2y/r - 3x/r = r; multiply each side by r.
2y - 3x = r^2 = x^2 + y^2.
Now, we have 2y - 3x = x^2 + y^2. We could be done, but we probably want to make this look like something, so let's complete the square to get it into the standard conic equation form.
x^2 + 3x + y^2 - 2y = 0.
So, to complete the square for x, we add (b/2)^2 to each side; same for y, so:
x^2 + 3x + 9/4 + y^2 - 2y + 1 = 9/4 + 1.
Now, factor according to how we complete the square:
(x + 3/2)^2 + (y - 1)^2 = 13/4.
What you should see, now, is that, as the "x^2" and "y^2" terms are both multiplied by the same number (i.e. 1), then we have a circle with center at (-3/2, 1) and radius sqrt(13)/2. Remember that the standard form of a circle is (x - h)^2 + (y - h)^2 = R^2, where (h, k) is the center and R the radius.
(2) sin(T + pi/4) can be expanded by the trig. addition formula sin(A + B) = sinAcosB - cosAsinB. So, we simply get sqrt(2)/2(sinT + cos T). Now, if we multiply by r, we get sqrt(2)/2(y + x) = 6. We may manipulate this equation further, but you should see, by now, that we have a line. The slope-intercept form is y = -x + 6sqrt(2).