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Question: How do you convert \({y^2} = 9x\) to polar form....

How do you convert y2=9x{y^2} = 9x to polar form.

Explanation

Solution

The conversion from rectangular to polar form is-
x=rcos⁡θx = r\cos \theta
y=rsin⁡θy = r\sin \theta
Use trigonometric identities such as cos⁡θsin⁡θ=cot⁡θ\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta and 1sin⁡θ=csc⁡θ\dfrac{1}{{\sin \theta }} = \csc \theta

Complete step by step solution:
As per the question convert y2=9x{y^2} = 9x to polar form is:
The conversion from rectangular to polar form is:
x=rcos⁡θx = r\cos \theta
y=rsin⁡θy = r\sin \theta
Given equation,
y2=9x{y^2} = 9x
Put the value x=rcos⁡θx = r\cos \theta and y=rsin⁡θy = r\sin \theta in the above equation to convert it into polar form.
Therefore,
(rsin⁡θ)2=9(rcos⁡θ){\left( {r\sin \theta } \right)^2} = 9\left( {r\cos \theta } \right)
Here, (rsin⁡θ)2{\left( {r\sin \theta } \right)^2} will distribute the square after opening the bracket.
Therefore the modified equation will be,
r2sin⁡2θ=9rcos⁡θ{r^2}{\sin ^2}\theta = 9r\cos \theta
In the above equation 9rcos⁡θ9r\cos \theta is transferred to the right side and the sign will also change as it is shifted from right side to left side.
So,
r2sin⁡2θ−9rcos⁡θ=0{r^2}{\sin ^2}\theta - 9r\cos \theta = 0
Take the common term, from the above equation.
r(rsin⁡2θ−9cos⁡θ)=0r\left( {r{{\sin }^2}\theta - 9\cos \theta } \right) = 0
At this point either r=0r = 0 or rsin⁡2θ−9cos⁡θ=0r{\sin ^2}\theta - 9\cos \theta = 0
Let’s solve the second one to get a meaningful answer.
As, r=0r = 0
Therefore,
rsin⁡2θ−9cos⁡θ=0r{\sin ^2}\theta - 9\cos \theta = 0
Shift ′−9cos⁡θ′' - 9\cos \theta ' to right side
∴rsin⁡2θ=9cos⁡θ\therefore r{\sin ^2}\theta = 9\cos \theta
Write the value in terms of ′r′'r'
r=9cos⁡θsin⁡2θr = \dfrac{{9\cos \theta }}{{{{\sin }^2}\theta }}
Here, sin⁡2θ{\sin ^2}\theta can be written as sin⁡θ×sin⁡θ\sin \theta \times \sin \theta
Therefore,
r=9cos⁡θsin⁡θ.1sin⁡θr = \dfrac{{9\cos \theta }}{{\sin \theta }}.\dfrac{1}{{\sin \theta }}
As from the trigonometric identities.
cos⁡θsin⁡θ=cot⁡θ\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta and 1sin⁡θ=csc⁡θ\dfrac{1}{{\sin \theta }} = \csc \theta
After putting the above identities, the value of ′r′'r' will be r=9cot⁡θ.csc⁡θr = 9\cot \theta .\csc \theta

Hence, the polar form of y2=9x{y^2} = 9x is r=9cot⁡θ.csc⁡θr = 9\cot \theta .\csc \theta

Additional information:
The polar form of a complex number is another way to represent a complex number. The form z=a+biz = a + bi is called the rectangular coordinate form of complex numbers.
The horizontal axis is the real axis and the vertical axis is the imaginary axis. You find the real and complex component in terms of ′r′'r' and ′θ′'\theta ' where ′r′'r' is the length of the vector and ′θ′'\theta ' is the angle made with the real axis.

Note: If you change the place of any number the sign will also get changed.
As,
′+′' + ' Addition will convert into ′−′' - ' subtraction.
′−′' - ' Subtraction will convert into ′+′' + ' Addition
′×′' \times ' Multiplication will convert into ′÷′' \div ' division
′÷′' \div ' division will convert into ′×′' \times ' multiplication.
Use trigonometric identities for converting into polar form.
For example: cos⁡θsin⁡θ=cot⁡θ,1sin⁡θ=csc⁡θ\dfrac{{\cos \theta }}{{\sin \theta }} = \cot \theta ,\dfrac{1}{{\sin \theta }} = \csc \theta