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Q11E

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Found in: Page 535

### Essential Calculus: Early Transcendentals

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

# (a) Find the eccentricity, (b) identify the conic, (c) give an equation of the directrix, and (d) sketch the conic.$${\rm{r = }}\frac{{\rm{2}}}{{{\rm{3 + 3sin\theta }}}}$$

(a) Eccentricity $${\rm{e = 1}}$$.

(b) The conic is parabola.

(c) The directrix $${\rm{y = }}\frac{{\rm{2}}}{{\rm{3}}}$$.

See the step by step solution

## Step 1: Find the eccentricity

(a)

If the focus is origin then the Eccentricity $${\rm{e}} \ne {\rm{1}}$$

Directrix $${\rm{x}} = \pm {\rm{d}}$$

$${\rm{r}} = \frac{{{\rm{ed}}}}{{{\rm{1}} \pm {\rm{ecos\theta }}}}$$

If the focus is origin Eccentricity $${\rm{e}} \ne {\rm{1}}$$

Directrix $${\rm{y}} = {\rm{ \pm d}}$$

$$r = \frac{{ed}}{{1 \pm e\sin \theta }}$$

$${\rm{r}} = \frac{{\rm{2}}}{{{\rm{3 + 3sin\theta }}}}$$

Numerator and denominator divided by 3.

$${\rm{r}} = \frac{{{\raise0.7ex\hbox{2} \!\mathord{\left/ {\vphantom {2 3}}\right.\kern-\nulldelimiterspace} \!\lower0.7ex\hbox{3}}}}{{{\rm{1 + sin\theta }}}}$$

We can read the coefficient of sin $${\rm{\theta }}$$ as the eccentricity, which is $${\rm{e = 1}}$$, because the constant term in the denominator is now $${\rm{1}}$$.

Thus, Eccentricity $${\rm{e = 1}}$$.

## Step 2: Identify the conic.

(b)

It is known that,

Eccentricity of ellipse is less than $${\rm{1}}$$.

Eccentricity of parabola is equal to $${\rm{1}}$$.

Eccentricity of hyperbola is greater than $${\rm{1}}$$.

Here, Eccentricity is $${\rm{e}} = {\rm{1}}$$.

The given conic is a parabola.

## Step 3: Equation of the directrix.

(c)

When divide the right term of the given equation by 3.

\begin{aligned}{l}{\rm{r}} = \frac{{\rm{2}}}{{{\rm{3 + 3sin\theta }}}}\\{\rm{r}} = \frac{{\frac{{\rm{2}}}{{\rm{3}}}}}{{{\rm{1 + sin\theta }}}}\end{aligned}

The equation of form

$${\rm{r}} = \frac{{{\rm{ed}}}}{{{\rm{1 + esin\theta }}}}$$

It is clear that, e=1 is the denominator.

$${\rm{ed}} = \frac{{\rm{2}}}{{\rm{3}}}$$is the numerator.

Substituting e in $${\rm{2nd }}$$equation,

\begin{aligned}{c}\left( {\rm{1}} \right){\rm{d}} = \frac{{\rm{2}}}{{\rm{3}}}\\{\rm{d}} = \frac{{\rm{2}}}{{\rm{3}}}\end{aligned}

The directrix $${\rm{y}} = \frac{{\rm{2}}}{{\rm{3}}}$$.

## Step 4: Graph of the conic.

(d)

The graph of conic is shown below: