An ellipse has the equation $$\frac{(x-\tfrac{1}{3})^2}{\tfrac{4}{9}}+\frac{y^2}{\tfrac{1}{3}}=1\;,$$

with focal points $(0,0)$ and $(2/3,0)$.

If a point P on the ellipse has a distance $1/2$ from the origin, what is its distance from the other focus? I initially thought it was to simply use $d_1+d_2=2a$ but found out this was wrong. My second attempt was to try and work out coordinates for P where I got $(0,1/2)$ and $(0,−1/2)$ therefore giving a distance of $5/6$ from the other focus. However I do not think this is right as I got complex solutions along the way which I just ignored.. wondered if there was another way?

  • 1
    $\begingroup$ We have $PF_1+PF_2=2a=2(2/3)=4/3$ where $F_1(0,0),F_2(2/3,0)$. So, the answer is $4/3-1/2=5/6$. $\endgroup$
    – mathlove
    Nov 18, 2016 at 15:01
  • $\begingroup$ You should say that it is the same ellipse you had already considered some days ago in this question. This information can be beneficial to people in order to spare them time. $\endgroup$
    – Jean Marie
    Nov 19, 2016 at 11:29

1 Answer 1


I'm assuming there is some educational reason why you're not supposed to use the formula $d_1+d_2=2a$. That formula gives the correct result for good reasons.

The method in which you solve two simultaneous quadratic equations is legitimate; after eliminating $y$, we can show that $x(x-2)=0$. If we allow complex solutions, this leads to four solutions; but since the points on the ellipse have real coordinates, it is perfectly legitimate to discard the complex solutions. The fact that $x$ and $y$ must be real is simply a constraint on the solution space; you can even add it to your system of equations as follows: \begin{gather} \tfrac94(x-x_0)^2 + 3y^2 = 1,\\ x^2 + y^2 = 1,\\ \Im (x) = 0,\\ \Im (y) = 0. \end{gather} The last two equations simply say that $x$ and $y$ have zero imaginary parts.

Another technique is to convert the equation of the ellipse to polar coordinates. The ellipse equation you were given was in the general form $$ \frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1 $$ for positive $a$ and $b$, where in this particular case $x_0=\frac13$, $y_0=0$, $a=\frac23$, and $b=\frac{1}{\sqrt3}$. Since $\frac23 > \frac{1}{\sqrt3}$, the length of the semi-major axis is $a$. In polar coordinates, the general equation for an ellipse with semi-major axis $a$, semi-minor axis $b$, one focus at $(0,0)$, and the other focus on the positive $x$-axis, like this ellipse, is $$ r = \frac{a(1 - e^2)}{1 - e\cos\theta} $$ where $e = \sqrt{1 - \left(\frac ba\right)^2}$. Plugging in $a=\frac23$ and $b=\frac{1}{\sqrt3}$, we find that $e = \sqrt{1 - \frac34} = \frac12$, so the equation of this particular ellipse comes out to $$ r = \frac{\frac23\left(1 - \left(\frac12\right)^2\right)} {1 - \frac12\cos\theta} = \frac{1}{2 - \cos\theta}. $$ But since $r$ in this equation is simply the distance from $(0,0)$ to a point on the ellipse at angle $\theta$ counterclockwise from the positive $x$-axis, and the desired point $P$ is at distance $\frac12$ from $(0,0)$, the polar coordinates of $P$ must satisfy $$ \frac12 = \frac{1}{2 - \cos\theta}, $$ from which we can deduce that $\cos\theta = 0$, and therefore $\sin\theta = \pm 1$. Converting the polar coordinates back into $x,y$ coordinates gives us \begin{align} x &= r \cos\theta = 0,\\ y &= r \sin\theta = \pm\frac12. \end{align} Therefore $P = \left(0,\frac12\right)$ or $P = \left(0,-\frac12\right)$.


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