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Q11E

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Essential Calculus: Early Transcendentals
Found in: Page 535
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Illustration

Short Answer

(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 by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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:

Most popular questions for Math Textbooks

Find the area of the region that is bounded by the given curve and lies in the specified sector.

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To determine the area enclosed by the one loop of the curve \({r^2} = \sin 2\theta \).

Find the area of the region that is bounded by the given curve and lies in the specified sector.

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{\vphantom {{\rm{\pi }} {\rm{3}}}} \right.

\kern-\nulldelimiterspace} {\rm{3}}}\)

Sketch the curve \({\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \right)^{\rm{3}}}{\rm{ = 4}}{{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{2}}}{\rm{. }}\)

Show that the polar curve\({\rm{r = 4 + 2sec\theta }}\) (called conchoids) has the line\({\rm{x = 2}}\) as a vertical asymptote by showing that \(\mathop {{\rm{lim}}}\limits_{{\rm{r}} \to {\rm{ \pm x}}} {\rm{x = 2}}\) . Use this fact to help sketch the conchoids.
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