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Question

Question: How do you convert \(r = - 2\cos ec\theta \) into Cartesian form ?...

How do you convert r=2cosecθr = - 2\cos ec\theta into Cartesian form ?

Explanation

Solution

In the given question, we are required to find an equation involving polar coordinates into an equation with Cartesian coordinates. Now, to convert polar coordinates into Cartesian coordinates, we have to replace x as rcosθr\cos \theta and replace y as rsinθr\sin \theta . Using these replacements, we can convert the polar coordinates into Cartesian coordinates with ease.

Complete step by step solution:
So, we have, r=2cosecθr = - 2\cos ec\theta
Firstly, we take all the parameters to the left side of the equation and all the constants to the right side of the equation.
rcosecθ=2\Rightarrow \dfrac{r}{{\cos ec\theta }} = - 2
Now, we know that cosecθ=1sinθ\cos ec\theta = \dfrac{1}{{\sin \theta }}. So, we also have sinθ=1cosecθ\sin \theta = \dfrac{1}{{\cos ec\theta }}. Hence, we get,
rsinθ=2\Rightarrow r\sin \theta = - 2
Now, we can replace rsinθr\sin \theta with y so as to convert the equation given in polar coordinates into Cartesian coordinates. Now, simplifying further, we get,
y=2\Rightarrow y = - 2
So, the Cartesian form of the equation given to us in the polar coordinates as r=2cosecθr = - 2\cos ec\theta is y=2y = - 2.

Note: For converting the polar coordinates into Cartesian coordinates, we replace the parameters of the polar coordinate system into Cartesian coordinate systems. So, we have x as rcosθr\cos \theta and replace y as rsinθr\sin \theta . Also, when we square and add both the equations, we obtain the relation r2sin2θ+r2cos2θ=y2+x2{r^2}{\sin ^2}\theta + {r^2}{\cos ^2}\theta = {y^2} + {x^2} which can be further simplified as r2=y2+x2{r^2} = {y^2} + {x^2} as we know that sin2θ+cos2θ=1{\sin ^2}\theta + {\cos ^2}\theta = 1 as a trigonometric identity. Therefore, the result r2=y2+x2{r^2} = {y^2} + {x^2} can be used directly for solving such questions if required.