Problem 76 - Algebra and Trigonometry

Question

Show that each statement is true by converting the given polar equation to a rectangular equation. Show that the graph of \(r=a \csc \theta\) is a horizontal line \(a\) units above the \(x\) -axis if \(a>0\) and \(|a|\) units below the \(x\) -axis if \(a<0\)

Step-by-Step Solution

Answer

The rectangular form of the polar equation \(r=a \csc \theta\) is \(r = \sqrt{x^2 + a^2}\). This implies that \(r\) is a constant, equal to \(a\). Therefore, the graph of the equation will be a horizontal line a units above the x-axis if \(a > 0\) and |a| units below the x-axis if \(a < 0\).

Step 1: Convert r to Rectangular Coordinates

The equation in polar coordinates is given as \(r = a \csc \theta\). The csc function, or cosecant, is defined as the reciprocal of the sine function, so you can rewrite \(r = a \csc \theta\) as \(r = a \sin^{-1} \theta\). Then, apply the Pythagorean identity to express sin θ in terms of 'r': \(\sin \theta = \frac{y}{r}\). Therefore, \(r = a \Big(\frac{y}{r}\Big)^{-1}\), and simplifying gives \(r = \frac{a}{y}\). On inverting both sides to get y in terms of r, we obtain \(y = \frac{a}{r}\)

Step 2: Substitute r in Rectangular Form

Next, substitute this newly expressed y into the Pythagorean identity for rectangular coordinates: \(r^2 = x^2 + y^2\), which gives \(r^2 = x^2 + (\frac{a}{r})^2\), and simplifying gives \(r^2 = x^2 + a^2\). Taking the square root of both sides gives \(r = \sqrt{x^2 + a^2}\).

Step 3: Conclude and Resolve the Statement

Since here we started with \(r = a \csc \theta\), and through converting and substitution got \(r = \sqrt{x^2 + a^2}\), it's clear \(r\) must be a constant equal to \(a\), as the square root of a square term remains constant. So for \(a > 0\), the graph will be a horizontal line a units above the x-axis. If \(a < 0\), then \(|a|\) (the absolute value of a) will be a positive value, so the graph will be |a| units below the x-axis.