Hello I am having some trouble understanding how to graph some conic equations, and especially what the value of eccentricity is given.
for example, one of the questions given was
Draw an ellipse with focus $(1,0)$ and directrix at $x=0$ with eccentricity $$\epsilon=\frac{1}{2}$$..
Because $$\epsilon=\frac{PF}{PD}$$ where P is any point on the ellipse, F a focus, and D the directrix, we have that $$2PF=PD$$
Now the professor said the following, let $P_{1}$ be a vertex on the left of the ellipse, so that x is a point on the ellipse.
Then if x is the distance from the origin to $P_{1}$ then the distance from P to the focus is $1-x$
so $2(1-x)=x$
$2-2x=x$
$2=3x$
$x=\frac{2}{3}$
So that seems to make sense and gives the placment of the first vertex,
Now im not sure I understand the next part and the general method that follows.
Now if we let the distance from the origin to the second vertex be x , then the distance from this point to the first focus is (x-1)
so $2(x-1)=x$
$2x-2=x$
$x=2$ But im not sure what this would give me. How can I find the right vertex from that and the second foci?
And in general just to provide another similar example, say we were doing the same but with focus $(5,0)$ and a directrix at $x=0$ with $\epsilon=\frac{1}{5}$
then would it be the same? we would just do $5(5-x)-x$ ie $x=\frac{25}{6}$ and the opposite to get $x=\frac{4}{5}$? but what does this give us?
And lastly, if we have a $\epsilon$ larger then zero, I would get vales for x to be $x=\frac{5}{6}$ and $x=\frac{-4}{5}$ but since the directrix is in the middle is this now a hyperbola?
Sorry for the long question, hope someone can take some time to help me understand.
I am still having some difficulties with this because I am not sure if my thoughts are correct . Especially in regard to the other examples I gave
Thank you
By this method you get the two vertices (beware $x=2/3$ and not $3/2$ in your first example). Their midpoint is the center of the ellipse, the second focus is the symmetric of the first one with respect to the center.