8.5 Conic sections in polar coordinates  (Page 3/8)

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 2, which is $\frac{1}{2}.$

Because $e=\frac{3}{2},e>1,$ so we will graph a hyperbola    with a focus at the origin. The function has a $\sin \theta$ term and there is a subtraction sign in the denominator, so the directrix is $y=-p.$

The directrix is $y=-\frac{8}{3}.$

Plotting a few key points as in [link] will enable us to see the vertices. See [link] .

First, we rewrite the conic in standard form by multiplying the numerator and denominator by the reciprocal of 5, which is $\frac{1}{5}.$

Because $e=\frac{4}{5},e<1,$ so we will graph an ellipse    with a focus at the origin. The function has a $\text{cos}\theta ,$ and there is a subtraction sign in the denominator, so the directrix    is $x=-p.$

The directrix is $x=-\frac{5}{2}.$

Plotting a few key points as in [link] will enable us to see the vertices. See [link] .

The directrix is $y=-p,$ so we know the trigonometric function in the denominator is sine.

Because $y=\mathrm{-2},\mathrm{–2}<0,$ so we know there is a subtraction sign in the denominator. We use the standard form of

and $e=3$ and $\left|\mathrm{-2}\right|=2=p.$

Therefore,

Because the directrix is $x=p,$ we know the function in the denominator is cosine. Because $x=4,4>0,$ so we know there is an addition sign in the denominator. We use the standard form of

and $e=\frac{3}{5}$ and $\left|4\right|=4=p.$

Therefore,