1
$\begingroup$

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

$\endgroup$
1

3 Answers 3

Reset to default
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$ \epsilon = \sqrt (1-a^2/b^2)=\sqrt 7/4 $$

$\endgroup$
2
  • $\begingroup$ Thanks. Can you eleaborate? For example, what formula are you using for $\epsilon$ ? $\endgroup$
    – Quality
    Commented Oct 11, 2015 at 19:05
  • $\begingroup$ @Quality Please visit this site Formulae $\endgroup$
    – Sathyaram
    Commented Oct 12, 2015 at 7:05
0
$\begingroup$

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:

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

$\endgroup$
2
  • $\begingroup$ Thank you, how do you get that $$PF=4-\sqrt{7}$$, because isnt the x distance from the vertex (-3,0) to the focus is 3, and the height is sqrt(7) so shouldnt the distance be d^2=3^2+7 ie d=4? $\endgroup$
    – Quality
    Commented Oct 11, 2015 at 19:20
  • $\begingroup$ If above edit is not clear I would upload a picture. Forget (-3,0), the directrix can be horizontal also as it is in this case. $\endgroup$
    – Narasimham
    Commented Oct 12, 2015 at 5:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .