Question

(a) Find the eccentricity, (b) identify the conic, (c) give an equation of the directrix, and (d) sketch the conic \({\rm{r = }}\frac{{\rm{5}}}{{{\rm{2 - 2sin\theta }}}}\).

Short answer

\(d = \frac{5}{2}\)

Step-by-step solution

Eccentricity Derivation.

(a)

The polar curve can be expressed in the following way:

\({\rm{r = }}\frac{{\rm{5}}}{{{\rm{2 - 2sin\theta }}}}\)

\({\rm{r = }}\frac{{\left( {\frac{{\rm{5}}}{{\rm{2}}}} \right)}}{{{\rm{1 - sin\theta }}}}\)

The most Common polar equation is the polar equation in its conventional form, gives

\({\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 - esin\theta }}}}\)

When the aforementioned equation is compared to the gives polar equation, gives

\({\rm{ed = }}\frac{{\rm{5}}}{{\rm{2}}}\)

\({\rm{e = 1}}\)

\({\rm{d = }}\frac{{\rm{5}}}{{\rm{2}}}\)

Eccencitity \({\rm{e = 1}}\).

Conic Derivation.

The polar curve can be expressed in the following way:

\({\rm{r = }}\frac{{\rm{5}}}{{{\rm{2 - 2sin\theta }}}}\)

\({\rm{r = }}\frac{{\left( {\frac{{\rm{5}}}{{\rm{2}}}} \right)}}{{{\rm{1 - sin\theta }}}}\)

The most Common polar equation is the polar equation in its conventional form, gives

\({\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 - esin\theta }}}}\)

When the aforementioned equation is compared to the gives polar equation, gives

\({\rm{ed = }}\frac{{\rm{5}}}{{\rm{2}}}\)

\({\rm{e = 1}}\)

\({\rm{d = }}\frac{{\rm{5}}}{{\rm{2}}}\)

Because of the quirkiness,\({\rm{e = 1}}\)Consequently, the conic is a parabola.

Since\({\rm{ e = 1}}\) is a Parabolic conic.

Directix Equation.

The polar curve can be expressed in the following way:

\(r = \frac{5}{{2 - 2\sin \theta }}\)

\(r = \frac{{\left( {\frac{5}{2}} \right)}}{{1 - \sin \theta }}\)

The most Common polar equation is the polar equation in its conventional form, gives

\(r = \frac{{ed}}{{1 - e\sin \theta }}\)

When the aforementioned equation is compared to the gives polar equation, gives

\(\begin{aligned}{l}ed = \frac{5}{2}\\{\rm{e = 1}}\\d = \frac{5}{2}\end{aligned}\)

Directix Equation, \(y = - \frac{5}{2}\)

Conic.

The polar curve can be expressed in the following way:

\(\begin{aligned}{l}{\rm{r = }}\frac{{\rm{5}}}{{{\rm{2 - 2sin\theta }}}}\\{\rm{r = }}\frac{{\left( {\frac{{\rm{5}}}{{\rm{2}}}} \right)}}{{{\rm{1 - sin\theta }}}}\end{aligned}\)

The most Common polar equation is the polar equation in its conventional form, gives

\({\rm{r = }}\frac{{{\rm{ed}}}}{{{\rm{1 - esin\theta }}}}\)

When the aforementioned equation is compared to the gives polar equation, gives

\(\begin{aligned}{l}{\rm{ed = }}\frac{{\rm{5}}}{{\rm{2}}}\\{\rm{e = 1}}\\{\rm{d = }}\frac{{\rm{5}}}{{\rm{2}}}\end{aligned}\)

Graph.