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Found in: Page 535

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# To determine,a) The eccentricity of the equation, $$r = \frac{4}{{5 - 4\sin \theta }}$$.b) To identify the shape of conic.c) The equation of the directrix.d) To sketch the conic.

a) The eccentricity of the equation, $$r = \frac{4}{{5 - 4\sin \theta }}$$ is $$e = \frac{4}{5}$$.

b) The conic shape is Ellipse.

c) The equation of directrix is $$y = - 1$$.

d) It is observed that the vertices of the ellipse are $$\left( {4,\frac{\pi }{2}} \right)$$ and $$\left( {\frac{4}{9},\frac{{3\pi }}{2}} \right)$$.

See the step by step solution

## Step 1: Given data

The given polar equation is $$r = \frac{4}{{5 - 4\sin \theta }}$$.

## Step 2: Theorem of Polar equation

A polar equation of the form $$r = \frac{{ed}}{{{l_{ \pm e}}\sin \theta }}$$ or $$r = \frac{{ed}}{{{1_{ \pm e}}\cos \theta }}$$ represents a conic section with eccentricity $$e$$.

The conic is an ellipse if $$e < 1$$, a parabola if $$e = 1$$, or a hyperbola if $$e > 1$$.

## Step 3: Calculation for the eccentricity of the equation

a)

Obtain the eccentricity of the given equation.

The polar equation for the given equation will be either $$r = \frac{{ed}}{{{1_{ \pm e}}\sin \theta }}$$ or $$r = \frac{{ed}}{{{1_{ \pm e}}\cos \theta }}$$.

Since the given polar equation is $$r = \frac{4}{{5 - 4\sin \theta }}$$, divide the given equation by $$5$$.

\begin{aligned}{l}r = \frac{4}{{5 - 4\sin \theta }}\\r = \frac{{\frac{4}{3}}}{{\frac{3}{5} - \frac{4}{5}\sin \theta }}\\r = \frac{{\frac{4}{3}}}{{1 - \frac{4}{3}\sin \theta }}\end{aligned}

From the above-mentioned theorem, it is clear that $$e = \frac{4}{5}$$.

Therefore, the eccentricity of the equation, $$r = \frac{4}{{5 - 4\sin \theta }}$$ is $$e = \frac{4}{5}$$.

## Step 4: Identify the shape of conic

b)

From the above-mentioned theorem, there are some conditions to identify the conic. They are, if eccentricity is less than one, then the conic is Ellipse.

If eccentricity is equal to one, then the conic is Parabola.

If eccentricity is greater than one, then the conic is Hyperbola.

From part (a), the calculated eccentricity is $$\frac{4}{5}$$, that is less than one.

Hence, the conic is Ellipse.

## Step 5: Calculation of the equation of the directrix

c)

Since the eccentricity $$e = \frac{4}{5}$$, the polar equation can be rewrite as $$r = \frac{{\frac{4}{5}}}{{1 - \frac{4}{5}\sin \theta }}$$.

Compare the equation with the standard equation $$r = \frac{{ed}}{{{l_{ \pm e}}\sin \theta }}$$.

Equate the numerator as shown below.

\begin{aligned}{c}ed = \frac{4}{5}\\\frac{4}{5}d = \frac{4}{5}\\d = 1\end{aligned}

That is, $$d = 1$$.

Since the term $$- e\sin \theta$$ appears in the denominator, obviously, the directrix must be below the focus at the origin.

Therefore, the equation of directrix is $$y = - 1$$.

## Step 6: Calculation of the values of $$r$$for different $$\theta$$

d)

The polar co-ordinates are $$x = r\cos \theta$$ and $$y = r\sin \theta$$.

Substitute 0 for $$\theta$$ in $$r = \frac{4}{{5 - 4\sin \theta }}$$.

\begin{aligned}{l}r = \frac{4}{{5 - 4\sin (0)}}\\r = \frac{4}{5}\end{aligned}

Similarly calculate the values of $$r$$ for different values of $$\theta$$.

Tabulate the calculated values as shown below in the table.

## Step 7: Sketch the diagram of an ellipse

Sketch the ellipse $$r = \frac{4}{{5 - 4\sin \theta }}$$ as shown below in the figure.

From figure, it is observed that the vertices of the ellipse are $$\left( {4,\frac{\pi }{2}} \right)$$ and $$\left( {\frac{4}{9},\frac{{3\pi }}{2}} \right)$$.