Chapter 9: Q. 62 (page 732)

Prove that the graph of the equation: $r = k s e c θ$ , $- \frac{π}{2} < θ < \frac{π}{2}$ is a vertical line for any value of $k \neq 0$

Short Answer

Expert verified

The equation in rectangular form is $x = k$

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:

$r = k s e c θ$

$- \frac{π}{2} < θ < \frac{π}{2}$ and $k \neq 0$

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

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

$x = r \cos θ y = r \sin θ r = \sqrt{x^{2} + y^{2}}$

So,

$\cos θ = \frac{x}{r} s e c θ = \frac{r}{x}$

Now substitute expression of $s e c θ$ and $r$ into given equation.

$r = k s e c θ r = k (\frac{r}{x}) x = k$

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

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Prove that the graph of the equation: $r = k s e c θ$ , $- \frac{π}{2} < θ < \frac{π}{2}$ is a vertical line for any value of $k \neq 0$

Short Answer

Step by step solution

Step 1. Given information:

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

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Prove that the graph of the equation: r=ksecθ, -π2<θ<π2is a vertical line for any value of k≠0

Short Answer

Step by step solution

Step 1. Given information:

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

One App. One Place for Learning.

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Prove that the graph of the equation: $r = k s e c θ$ , $- \frac{π}{2} < θ < \frac{π}{2}$ is a vertical line for any value of $k \neq 0$