Solveeit Logo
Login

Question

Question: How do you convert \( 4{{x}^{2}}y=1 \) into polar form?...

How do you convert 4x2y=14{{x}^{2}}y=1 into polar form?

Explanation

Solution

Hint : We will try to convert to polar form by determining the modulus form and the identities like x=rcos⁡θ;y=rsin⁡θ;x2+y2=r2x=r\cos \theta ;y=r\sin \theta ;{{x}^{2}}+{{y}^{2}}={{r}^{2}} . We explain the number of ways the position of a point or equation can be expressed in different forms. We also explain the ways the representation works for polar and cartesian form. Then we convert the given equation into polar form using the relations.

Complete step-by-step answer :
There are always two ways to represent any point equation in our general 2-D and 3-D surfaces. One being the polar form and other one being the cartesian form. The other name of the cartesian form is rectangular form.
In case of polar form, we use the distance and the angle from the origin to get the position of the point or curve.
The given equation 4x2y=14{{x}^{2}}y=1 is a representation of the cartesian form.
In case of polar form r represents the distance and θ\theta represents the angle.
In case of rectangular form, we use the coordinates from the origin to get the position of the point or curve. For two dimensional things we have X-Y and for three dimensional things we have X-Y-Z. We take the perpendicular distances from the axes.
We need to convert the given equation 4x2y=14{{x}^{2}}y=1 into the polar form.
The relation between these two forms in two-dimensional is
x=rcos⁡θ;y=rsin⁡θ;x2+y2=r2x=r\cos \theta ;y=r\sin \theta ;{{x}^{2}}+{{y}^{2}}={{r}^{2}} .
Putting the values, we get
4x2y=1 ⇒4(rcos⁡θ)2(rsin⁡θ)=1 ⇒4r3cos⁡2θ=1sin⁡θ=csc⁡θ  4{{x}^{2}}y=1 \\\ \Rightarrow 4{{\left( r\cos \theta \right)}^{2}}\left( r\sin \theta \right)=1 \\\ \Rightarrow 4{{r}^{3}}{{\cos }^{2}}\theta =\dfrac{1}{\sin \theta }=\csc \theta \\\
The polar form of 4x2y=14{{x}^{2}}y=1 is 4r3cos⁡2θ=csc⁡θ4{{r}^{3}}{{\cos }^{2}}\theta =\csc \theta .
So, the correct answer is “ 4r3cos⁡2θ=csc⁡θ4{{r}^{3}}{{\cos }^{2}}\theta =\csc \theta ”.

Note : In case of points for cartesian form we use x and y coordinates as (x,y)\left( x,y \right) to express their position in the cartesian plane. The distance from origin is r=x2+y2r=\sqrt{{{x}^{2}}+{{y}^{2}}} . This r represents the distance in polar form.