Write in cartesian form:
r=3tan(theta)sec(theta)
im working with this function here, trying to get an rcos(theta) and/or rsin(theta)
so far, i am stumped.
what i have so far:
r^2=3r(tan(theta)sec(theta))
(x^2+y^2)=?????
sin=y/r
cos=x/r
3x^2=y
To convert the polar equation r = 3tan(theta)sec(theta) into Cartesian form, we can use the following relationships for tangent and secant:
tan(theta) = y / x
sec(theta) = r / x
Substituting these expressions into the equation r = 3tan(theta)sec(theta), we get:
r = 3(y / x)(r / x)
Next, we can multiply both sides by x to eliminate the denominators:
r * x = 3y * r
Now, we can substitute r with sqrt(x^2 + y^2) to get the equation in terms of x and y:
sqrt(x^2 + y^2) * x = 3y * sqrt(x^2 + y^2)
Squaring both sides of the equation, we get:
(x^2 + y^2) * x^2 = (3y)^2 * (x^2 + y^2)
Expanding and rearranging terms, we have:
x^4 + 2x^2y^2 + y^4 - 9x^2y^2 = 0
This equation represents the Cartesian form of the given polar equation r = 3tan(theta)sec(theta).
To write the equation r = 3tan(theta)sec(theta) in Cartesian form, we need to use the trigonometric identities to express tan(theta) and sec(theta) in terms of x and y.
Start with the identity: tan(theta) = y/x. Rearranging this equation, we get x = y/tan(theta).
Next, we use the identity: sec(theta) = 1/cos(theta). Rearranging this equation, we get cos(theta) = 1/sec(theta).
Now, let's substitute these identities back into the initial equation:
r = 3tan(theta)sec(theta)
r = 3(y/tan(theta))(1/sec(theta))
r = 3(y/x)(1/(1/cos(theta)))
r = 3(y/x)(cos(theta))
r = 3(y/x)(x/r)
r^2 = 3xy
Therefore, the Cartesian form of the polar equation r = 3tan(theta)sec(theta) is:
x^2 + y^2 = 3xy