A question about conic section (ellipse).

Question

I am asked to solve the problem what is the center of the ellipse with vertex $V_1=(1,3)$ and focus $F_1=(1,0)$ and eccentricity $e=1/2$.

My answer is due to the following analysis and computation:

Based from the positions of the given vertex and focus the major axis of my ellipse is vertical. And the center of the ellipse is located at $(1,k)$ where $k$ is a constant yet to determine.

Using $e=1/2=c/a$, this means that $2c=a$.

I know that the distance from the center $C$ and $V_1$ is $a$ so I have:

$d_{CV_1}=a=\sqrt{(1-1)^2+(3-k)^2}=3-k$.

also the distance from the center and the focus is $c$ so I have:

$d_{CF_1}=c=\sqrt{(1-1)^2+(0-k)^2}=k$.

Using the last two equations and the equation $c=2a$ it is easy to see that $k=1$. My conclusion is the center $C$ is located at $(1,1)$. However, graphing the points reveal that $C$ is located between $V_1$ and $F_1$ which is unusual for me.

My question is am I correct on my answer or something went wrong in the solving process. Thanks a lot.

Answer 1

Finding the solution

According to Wikipedia you have

$$ e = c/a $$

where $a$ is the semi-major axis, i.e. the distance between the center and either vertex, and $c$ (called $f$ on Wikipedia) is the focal distance, i.e. the distance between the center and either focus. Since you know

$$d_{F_1V_1} = a - c = 3$$

you can solve for $a$:

\begin{align*} e &= \frac ca \\ \frac12 &= \frac{a-3}{a} \\ a &= 2a-6 \\ a &= 6 \end{align*}

So the other vertex must be at distance $2a=12$ from the first, i.e. at $V_2=(1,-9)$, and the center would be at distance $a$ from the first focus, i.e. $C=(1,-3)$.

Comparing with your result

This would imply $k=-3$, in contrast to the $k=1$ you had. So where did you go wrong? The problem lies here:

$$d_{CF_1}=c=\sqrt{(1-1)^2+(0-k)^2}=k$$

There you incorrectly simplified $\sqrt{(0-k)^2}=k$ which should have been $\sqrt{(0-k)^2}=\lvert k\rvert$. Continuing with that absolute value would be a bit painful, though. My approach above can avoid that by fixing some signs in advance. In particular, I assume $a>c$ which must hold in any ellipse (as opposed to a hyperbola). That way, I don't have to write $\lvert a-c\rvert=3$ and can avoid sign distinctions.

Answer 2

I can't pinpoint your error but here's how I do it: The directrix $D_1$ associated with $ F_1$ passes thru $P_1=(1,d_1)$ with $ d_1>0$. We have $$3=|V_1F_1|=e.|DV_1|=(1/2)|P_1|=(1/2) |d_1-3|=(d_1-3)/2.$$ Therefore $$d_1=9.$$ The other vertex $ V_2=(w_2,1)$ (with $w_2 <0$) satisfies $$-w_2=|w_2|=|V_2F_1|=e|V_2F_1|=(1/2)|V_2D|=(1/2)|9-w_2|=(9-w_2)/2$$. Therefore$$ w_2=-9.$$ The center C is the midpoint between $V_1=(1,3)$ and $V_2=(1,-9)$ ,so $C=(1,-3)$.