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Question

Question: How do you write the given equation \[{x^2} - {y^2} = 1\] into polar form?...

How do you write the given equation x2y2=1{x^2} - {y^2} = 1 into polar form?

Explanation

Solution

Here we will simply substitute the value of x=rcosθ,y=rsinθx = r\cos \theta ,y = r\sin \theta in the equation and solve it. Then after solving it we will write it in the simplified form which will be the polar form of the given equation.

Complete Step by Step Solution:
Given equation is x2y2=1{x^2} - {y^2} = 1.
So, to convert the given equation into the polar form we will substitute x=rcosθ,y=rsinθx = r\cos \theta ,y = r\sin \theta . Therefore the equation becomes
(rcosθ)2(rsinθ)2=1\Rightarrow {\left( {r\cos \theta } \right)^2} - {\left( {r\sin \theta } \right)^2} = 1
Now we will simply open the brackets and square the terms in the bracket. Therefore, we get
r2cos2θr2sin2θ=1\Rightarrow {r^2}{\cos ^2}\theta - {r^2}{\sin ^2}\theta = 1
Now we will take r2{r^2} common from the terms of the equation, we get
r2(cos2θsin2θ)=1\Rightarrow {r^2}\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) = 1
We know that cos2θsin2θ=cos2θ{\cos ^2}\theta - {\sin ^2}\theta = \cos 2\theta . Therefore, we will put this value in the above equation. Therefore, the equation becomes
r2(cos2θ)=1\Rightarrow {r^2}\left( {\cos 2\theta } \right) = 1
Now we will take the term cos2θ\cos 2\theta to the other side of the equation, we get
r2=1cos2θ\Rightarrow {r^2} = \dfrac{1}{{\cos 2\theta }}
We know that the reciprocal of the cos trigonometric function is equal to sec trigonometric function. Therefore the equation becomes
r2=sec2θ\Rightarrow {r^2} = \sec 2\theta

Hence the given equation (x3)2+y2=9{\left( {x - 3} \right)^2} + {y^2} = 9 into polar form can be written as r2=sec2θ{r^2} = \sec 2\theta .

Note:
Here we have to note that the ratio of the sinθ\sin \theta and cosθ\cos \theta is equal to the tanθ\tan \theta . Also the ratio of cosθ\cos \theta and sinθ\sin \theta is equal to cotθ\cot \theta and the reciprocal of the cos function is equal to sec function and reciprocal of the sin function is equal to the cosec function.
tanA=sinAcosA,secA=1cosA,cosecA=1sinA\tan A = \dfrac{{\sin A}}{{\cos A}},\sec A = \dfrac{1}{{\cos A}},\cos ecA = \dfrac{1}{{\sin A}}
We don’t have to confuse the polar coordinate system with the normal rectangular coordinate system. Polar coordinate system is the system in which the coordinates of a point is represented by the distance of that point from a reference point and by the angle from the reference plane.