Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a couple of questions regarding ellipses.

Get the equation of the ellips

  • With Foci $(\pm 3,0)$ and which goes through $(2,\sqrt{2})$. This one I didn't understand AT ALL. I need some explanation

Valentin gave an answer I originally deemed correct, however, you can see my objections in the comments of the answer.

share|cite|improve this question
Do you mean goes through $(2,\sqrt{2})$? – littleO Nov 9 '12 at 9:10
@littleO yes, I do mean that, I'll edit it – JohnPhteven Nov 9 '12 at 9:11
up vote 2 down vote accepted

The equation of an ellipse whose major and minor axes coincide with the Cartesian axes is \begin{equation} \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. \end{equation} The distance from the center to either focus is $f$ where \begin{equation} f^2 = a^2 - b^2. \end{equation} We are given that $f = 3$, from which we conclude that $b^2 = a^2 - 9$. The equation of our ellipse reduces to \begin{equation} \frac{x^2}{a^2} + \frac{y^2}{a^2 - 9} = 1. \end{equation} Now we plug in the point $(2,\sqrt{2})$ and obtain \begin{equation} \frac{4}{a^2} + \frac{2}{a^2 - 9} = 1. \end{equation} There are four values of $a$ that satisfy this equation, and we pick the one that is larger than $3$: \begin{equation} a = 2 \sqrt{3}. \end{equation}

How to solve for $a$ in more detail: \begin{align*} &\frac{4}{a^2} + \frac{2}{a^2 - 9} = 1 \\ \implies& 4 + \frac{2a^2}{a^2 - 9} = a^2 \\ \implies& 4(a^2 - 9) + 2a^2 = a^2(a^2 - 9) \\ \implies& a^4 - 15a^2 + 36 = 0 \\ \implies& (a^2 - 12)(a^2 - 3) = 0 \\ \implies& a = \pm 2\sqrt{3} \text{ or } a = \pm \sqrt{3}. \end{align*}

share|cite|improve this answer
How do you know which values satisfy the equation? Is there a handy way to get those values? – JohnPhteven Nov 9 '12 at 9:25
I just edited my answer to show that part in more detail. – littleO Nov 9 '12 at 9:30
Ok, thank you. Last question before I award you the deserved bounty, what is wrong about Valentin's method? – JohnPhteven Nov 9 '12 at 9:32
Valentin correctly set up the equation $a^4 - 15a^2 + 36 = 0$, but his next step did not follow and was an error. It would have been correct if he'd said $a^2 = 12$ or $a^2 = 3$. – littleO Nov 9 '12 at 9:35
The bounty is yours (in 23 hours) – JohnPhteven Nov 9 '12 at 9:36

We know, the sum of the distances from any point $P(h,k)$ on the ellipse to those two foci is constant and equal to the major axis $(2a)$.

As $(2,\sqrt2)$ lies on the ellipse, $2a=\sqrt{(3-2)^2+(0-\sqrt 2)^2}+\sqrt{(-3-2)^2+(0-\sqrt 2)^2}=\sqrt 3+3\sqrt3=4\sqrt3$

So, $\sqrt{(h-3)^2+(k-0)^2}+\sqrt{\{h-(-3)\}^2+(k-0)^2}=2a=4\sqrt 3$

or ,$\sqrt{(h-3)^2+(k-0)^2}=4\sqrt 3-\sqrt{\{h-(-3)\}^2+(k-0)^2}$

On squaring, $(h-3)^2+(k-0)^2=48+(h+3)^2+k^2-8\sqrt3\sqrt{(h+3)^2+k^2}$

or, $8\sqrt3\sqrt{(h+3)^2+k^2}=48+12h$

On squaring and simplification, $$h^2+4k^2=12\space or \space \frac {h^2}{12}+\frac{k^2}3=1$$

So, the locus of $P(h,k)$ is $$\frac {x^2}{12}+\frac{y^2}3=1$$


$a=2\sqrt 3$.

Here $ae=3,e=\frac3{2\sqrt3}=\frac{\sqrt3}2$

If $2b$ is the minor axis,$b^2=a^2(1-e^2)=(2\sqrt3)^2\left(1-\frac3 4\right)=3$

We know, the midpoint of the segment connecting the foci is the centre of the ellipse, so here the centre is $\frac{3-3}2,\frac{0+0}2$ i.e., $(0,0)$

Again, the major axis is the segment that contains both foci.

Here the equation of the major axis is $\frac{y-0}{x-3}=\frac{0-0}{\{3-(-3)\}}$ i.e, $y=0$ the X axis.

So, the required equation of the ellipse is $$\frac{(x-0)^2}{(2\sqrt3)^2}+\frac{(y-0)^2}{3}=1 $$

share|cite|improve this answer

Your first solution is correct, though in case of the ellipse $a^2=b^2+c^2$ not directly by Pythagoras theorem, but rather by definition of $b$.

It is this definition that you need to use in the second part where $c=3$, so

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ $$\frac{x^2}{a^2}+\frac{y^2}{a^2-9}=1$$

Now that you are given the point (assume it is $(2,\sqrt{2})$) through which the ellipse passes, substitute its coordinates in the equation and solve for $a$. $$\frac{4}{a^2}+\frac{2}{a^2-9}=1$$

share|cite|improve this answer
Oh, is it that simple? I'm going crazy. I have missed an entire semester, I am almost finished in doing 3 months of maths in 2 days.. sorry if you see some stupid questions once in a while – JohnPhteven Nov 2 '12 at 22:54
I actually discovered I don't know how to solve $\dfrac{4}{a^2} + \dfrac{2}{a^2-9} = 1$ ... can you show me? – JohnPhteven Nov 2 '12 at 23:12
But what is $b$? Is it $4$? But if it's $4$, than how can $a^2-9$ equal $b^2$? – JohnPhteven Nov 9 '12 at 8:40
How can $c$ be bigger than $a$? This is a wrong answer! – JohnPhteven Nov 9 '12 at 8:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.