How to find directrix of conic

Question

Hello I am having some confusion with conic sections.

For example, I am asked to find the directrix and eccentricity of the ellipse given by the formula

$$\frac{x^2}{9}+\frac{y^2}{16}=1$$

So, what I know is that this ellipse has vertices $(0,4)$ and $(0,-4)$

I also know it has points at $(-3,0)$ and $(3,0)$ as the solve the equation

I know that the foci are at $c=(0,\sqrt{7})$ and $-c=(0,-\sqrt{7})$

Now , if we let $\epsilon$=eccentricity then we would have $\frac{PF}{PD}=\epsilon$

the distance from the point $(-3,0)$ to $(0,\sqrt{7})$ is $4$,

But I do not know how to find the directrix,

I am looking for any help to understand, thanks

Answer 1

Since the major axis of this ellipse is vertical, the foci have coordinates (0,be), (0,-be), and directrices have equation y=b/e, y=-b/e. b is the length of the semi-major axis which is 4.

Answer 2

It is next only a small step next, proceeding from the definition of eccentricity.

EDIT1: Consider the points only on y-axis $ O,F,P,D $ bottom to top.

$$ PF=OP-OF=4-c= 4-\sqrt7$$

$$ \epsilon < 1 = \frac{c}{a}= \frac{\sqrt{7}}{4}$$ $$ \frac{PD}{PF}= \frac{1}{ \epsilon } $$

The above holds good for any point on the ellipse.From this particularly,the distance from end of major axis to the horizontal directrix $PD$ can be found out:

$$ PD= \frac{4(4 \sqrt{7} -7)}{7}.$$

Answer 3

Since x = 0 is the equation major axis i.e. (b>a) the formula for directrices is
$$ y=+b/\epsilon $$ and $$ y=-b/\epsilon $$ according to the equation $$ x^2/3^2 + y^2/4^2 = 1 $$ b = 4 and $$ \epsilon = \sqrt (1-a^2/b^2)=\sqrt 7/4 $$