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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.

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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

3 Answers 3

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*}

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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 $$\frac{(x-0)^2}{(2\sqrt3)^2}+\frac{(y-0)^2}{3}=1 $$

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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 $a$. $$\frac{4}{a^2}+\frac{2}{a^2-9}=1$$

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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

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