Question

Prove that the graph of the equation: $r=ksec\theta$, $-\frac{\mathrm{\pi }}{2}<\theta <\frac{\mathrm{\pi }}{2}$is a vertical line for any value of $k\ne 0$

Short answer

The equation in rectangular form is $x=k$

So

It has been proved that graph of given equation is a vertical line.

Step-by-step solution

Step 1. Given information:

The polar form of equation:

$r=ksec\theta$

$-\frac{\mathrm{\pi }}{2}<\theta <\frac{\mathrm{\pi }}{2}$and$k\ne 0$

Step 2. Convert polar form of equation into rectangular form.

As we know the rectangular form is converted into polar form by using.

$$\begin{gathered} x=r\cos \theta \\ y=r\sin \theta \\ r=\sqrt{x^2+y^2} \end{gathered}$$

So,

$$\begin{gathered} \cos \theta =\frac{x}{r} \\ sec\theta =\frac{r}{x} \end{gathered}$$

Now substitute expression of $sec\theta$and $r$into given equation.

$$\begin{gathered} r=ksec\theta \\ r=k\left(\frac{r}{x}\right) \\ x=k \end{gathered}$$

Thus the graph of the given equation is a vertical line, which is at a distance of$k$units from$y$axis.