Step by step solution to "The eccentricities of conic sections with one focus..."

Alyce Wilkinson

Answered question

2021-03-09

The eccentricities of conic sections with one focus at the origin and the directrix corresponding and sketch a graph:
$e = 2,$
$r = - 2 \sec θ .$

Neelam Wainwright

Skilled2021-03-10Added 102 answers

e = 2, r = - 2 \sec θ

Since

e > 1

, therefore the conic is a hyperbola.
Now consider the directrix,

r = - 2 \sec θ

\Rightarrow r = - \frac{2}{\cos θ}

\Rightarrow r \cos θ = - 2

\Rightarrow x = - 2

Now comparing whith

x = - p,

we get

p = 2

Therefore equation is,

r = \frac{e p}{1 - e \cos θ}

\Rightarrow r = \frac{2 \cdot 2}{1 - 2 \cos θ}

\Rightarrow r = \frac{4}{1 - 2 \cos θ}

Now the graph is,

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