Question

How do you convert \[r=6\sec \theta \] into Cartesian form?

Answer 1

Hint: Use the conversion $\sec \theta =\dfrac{1}{\cos \theta }$ and cross-multiply the terms. Substitute the value of $\cos \theta $ in terms of x and r using the relation: - \[x=r\cos \theta \]. Here, x denotes the x – coordinate in Cartesian form. Now, cancel the common terms to get the required equation in Cartesian form.

Complete step by step answer:
Now, we can see that the given relation is in polar form because it relates the radius vector (r) and angle (\[\theta \]). We need to change it into Cartesian form that means we have to obtain the relationship between x and y – coordinate.
Let us consider a point (P) with Cartesian coordinates P (x, y) and polar coordinates \[\left( r,\theta \right)\], then the relation between the two-coordinate system is given as: -
\[\Rightarrow x=r\cos \theta \] and \[y=r\sin \theta \]
Considering the first relation we have,
$\Rightarrow \cos \theta =\dfrac{x}{r}................\left( i \right)$
Now, let us come to the question. So, we have,
\[\Rightarrow r=6\sec \theta \]
Using the relation: $\sec \theta =\dfrac{1}{\cos \theta }$, we get,
\[\begin{align}
  & \Rightarrow r=6\times \dfrac{1}{\cos \theta } \\
 & \Rightarrow r=\dfrac{6}{\cos \theta } \\
\end{align}\]
By cross – multiplication we get,
\[\begin{align}
  & \Rightarrow r=6\times \dfrac{1}{\cos \theta } \\
 & \Rightarrow r\cos \theta =6 \\
\end{align}\]
Using equation (1) and cancelling the common factors we get,
$\begin{align}
  & \Rightarrow r\times \dfrac{x}{r}=6 \\
 & \Rightarrow x=6 \\
\end{align}$
Hence, the above relation represents the given equation in Cartesian form.

Note:
One may note that the relation x = 6 represents a straight line parallel to the y axis. Here, the coefficient of y is 0. You must remember the relationship between the Cartesian coordinates and the polar coordinates to solve the above question. Also, remember the important relation: - \[{{x}^{2}}+{{y}^{2}}={{r}^{2}}\] as it will be used further in the coordinate geometry of circles. Do not consider the relation \[y=r\sin \theta \] as it relates the y – coordinate and sine of the given angle and it is of no use here.