Question

How do you convert polar equations to cartesian equations given $r\sec \theta = 3$ ?

Answer 1

Before solving such types of questions, we must know that the position of a point or equation can be expressed in several ways. We should also know the way polar form and cartesian form works for solving this question. We are given a polar equation and we will convert this into the cartesian equation by using the relation $x = r\cos \theta$ and $y = r\sin \theta$ , so we get the equation in terms of x and y.

Complete step by step solution:
A cartesian point is of the form $(x,y)$ and a cartesian equation is expressed in terms of x and y where x is the distance from the y-axis and y is the distance from the x-axis. A polar equation is of the form $(r,\theta )$ , where r is the distance from the origin and $\theta$ is the distance between the line from the origin to the point and the x-axis. So, x, y and r form a right triangle and from that we get –
$x = r\cos \theta$ , $y = r\sin \theta$ and ${r^2} = {x^2} + {y^2}$
We are given a polar equation –
$r\sec \theta = 3 \\\ \Rightarrow r \times \dfrac{1}{{\cos \theta }} = 3 \\\ \Rightarrow r = 3\cos \theta \\\$
Multiplying both the sides of the above equation by r, we get –
${r^2} = 3r\cos \theta$
Now, replacing the values of the known quantities, we get –
${x^2} + {y^2} = 3x$
The above equation is a cartesian equation.
Hence, the polar equation $r\sec \theta = 3$ is written in the form of the cartesian equation as ${x^2} + {y^2} = 3x$ .

Note: The obtained equation is in cartesian form, it can be further solved as follows,
${x^2} - 3x + {y^2} = 0$
Adding $\dfrac{9}{4}$ on both sides, we get –
${x^2} + \dfrac{9}{4} - 3x + {y^2} = \dfrac{9}{4} \\\ \Rightarrow {x^2} + {(\dfrac{3}{2})^2} - 3x + {y^2} = \dfrac{9}{4} \\\ \Rightarrow {(x - \dfrac{3}{2})^2} + {y^2} = \dfrac{9}{4} \\\$
This is an equation of a circle, on comparing it with the standard equation of the circle ${(x - h)^2} + {(y - k)^2} = {r^2}$ , we see that the obtained equation is an equation of the circle with the centre at $(\dfrac{3}{2},0)$ and radius $\dfrac{3}{2}units$ .