How to convert y^2-8x-16=0 to polar form
Why did the equation go to the party dressed as "y^2-8x-16=0" and not "polar form"? Because it wanted to be a square, not a circle! Okay, let's help this equation transform into its polar costume. We need to use the formulas r = sqrt(x^2 + y^2) and tan(theta) = y/x.
First, let's replace y with r sin(theta) and x with r cos(theta). Now the equation becomes:
(r sin(theta))^2 - 8(r cos(theta)) - 16 = 0
Let's expand this equation:
r^2 sin^2(theta) - 8r cos(theta) - 16 = 0
Now, let's use the trigonometric identity sin^2(theta) + cos^2(theta) = 1. Rearranging this, we get:
sin^2(theta) = 1 - cos^2(theta)
Let's substitute this identity into the equation:
r^2 (1 - cos^2(theta)) - 8r cos(theta) - 16 = 0
Now, simplify the equation:
r^2 - r^2 cos^2(theta) - 8r cos(theta) - 16 = 0
To convert this equation fully into polar form, we leave it in terms of r and theta without any x or y terms.
To convert the equation y^2 - 8x - 16 = 0 to polar form, follow these steps:
Step 1: Rewrite the equation in terms of polar coordinates.
Replace x with r*cos(theta) and y with r*sin(theta).
r^2*sin(theta)^2 - 8r*cos(theta) - 16 = 0
Step 2: Simplify the equation.
r^2*sin^2(theta) - 8r*cos(theta) - 16 = 0
Step 3: Expand the equation further.
r^2*sin^2(theta) - 8r*cos(theta) - 16 = r^2*(1 - cos^2(theta)) - 8r*cos(theta) - 16
Step 4: Simplify the equation.
r^2 - r^2*cos^2(theta) - 8r*cos(theta) - 16 = 0
Step 5: Group the terms and factor out r^2.
(r^2 - 8r*cos(theta)) - r^2*cos^2(theta) - 16 = 0
r^2(1 - cos^2(theta)) - 8r*cos(theta) - 16 = 0
Step 6: Replace 1 - cos^2(theta) with sin^2(theta) using the identity sin^2(theta) + cos^2(theta) = 1.
r^2*sin^2(theta) - 8r*cos(theta) - 16 = 0
Step 7: Simplify the equation.
r^2*(sin^2(theta) - 8*cos(theta)) - 16 = 0
This is the polar form of the given equation, y^2 - 8x - 16 = 0.
To convert the equation y^2 - 8x - 16 = 0 to polar form, we can use the polar coordinate transformation equations:
x = r * cos(theta)
y = r * sin(theta)
Let's start by substituting these equations into the given equation:
(r * sin(theta))^2 - 8(r * cos(theta)) - 16 = 0
Simplifying further, we get:
r^2 * sin^2(theta) - 8r * cos(theta) - 16 = 0
Now, to further simplify this equation, we can use trigonometric identities. Specifically, we know that sin^2(theta) + cos^2(theta) = 1, so we can rewrite sin^2(theta) as 1 - cos^2(theta):
r^2 * (1 - cos^2(theta)) - 8r * cos(theta) - 16 = 0
Expanding the expression, we get:
r^2 - r^2 * cos^2(theta) - 8r * cos(theta) - 16 = 0
Now, we can rearrange the equation to isolate the r term on one side:
r^2 - 8r * cos(theta) + (-r^2 * cos^2(theta) - 16) = 0
Next, we can factor out the r term:
r^2 - 8r * cos(theta) + (-r^2 * cos^2(theta) - 16) = r(r - 8 * cos(theta)) + (-r^2 * cos^2(theta) - 16) = 0
Now, we can substitute the expression inside the parenthesis into a new variable, let's say C:
C = -r^2 * cos^2(theta) - 16
Our equation simplifies to:
r(C + r - 8 * cos(theta)) = 0
Now we have the equation in polar form. The equation represents a curve in polar coordinates, where r is the radius and theta is the angle.
y = rsinØ, x = rcosØ
y^2 - 8x - 16 = 0
r^2 sin^2 Ø - 8rcosØ - 16 = 0
Can somone check my work please
y = rsinØ, x = rcosØ
y^2 - 8x - 16 = 0
r^2 sin^2 Ø - 8rcosØ - 16 = 0
(rsin(Ø))^2 - 8rcos(Ø) = 16
r^2sin^2(Ø) = 16+8rcos(Ø)
divide by r on both sides:
rsin^2(Ø) = 16+8cos(Ø)
divide by sin^2(Ø) to get r by itself:
r = 16+8cos(Ø) / (sin^2(Ø))
r = 16+8cos(Ø)csc^2(Ø)