Proving the equation of an ellipse

Question

If $(c, 0)$ and $(-c, 0)$ are the foci of an ellipse, and the sum of the distance of any point on the ellipse with the foci is $2a$ I am asked to prove thath the equation of the ellipse is: $$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $$ where $b^2=a^2-c^2$.

I tried to first write the definition of the ellipse in mathematical terms: $$ \sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a $$ But when I tried to proceed from here I didn't know where should I go.

1

Square the last formula. – Sebastiano Aug 03 '20 at 07:43

score 2 · Answer 1 · 2020-08-03T08:00:48.337

2

Hint:

square the two members to get a form like $p+2\sqrt{pq}+q=4a^2$;
isolate the square root in a member and square again to get a polynomial equation;
simplify and rearrange.

Addendum:

The squaring of $4a^2-p-q$ generates the quartic terms $p^2+2pq+q^2$. After subtraction of $4pq$ from the other member, this makes $(p-q)^2$, allowing the polynomial to remain quadratic.

edited Aug 03 '20 at 08:00

answered Aug 03 '20 at 07:44

2

Very synthetic but very efficient. – Sebastiano Aug 03 '20 at 07:47
1

@Sebastiano: I spoiled my answer with an addendum that nearly doubles the length :-( – Aug 03 '20 at 07:52
It is true :) ahaha. – Sebastiano Aug 03 '20 at 07:53

Hrishabh Nayal · Accepted Answer · 2020-08-03T07:58:22.563

2

A slightly faster way

Hint:

Move $\sqrt{(x+c)^2+y^2}$ to other side of equality to give $\sqrt{(x-c)^2+y^2}=2a-\sqrt{(x+c)^2+y^2}$

Square both sides.

The calculation should simplify a lot as a lot of terms cancel after first step ,compared to simply squaring original equation.

Then move terms within roots to another side and square again.

The problem should be solved.

edited Aug 03 '20 at 07:58

answered Aug 03 '20 at 07:51

Hrishabh Nayal

692

3

Why this downvote ? The method is indeed attractive. – Aug 03 '20 at 07:58
I get to $a^2x^2+a^2c^2+a^2y^2=a^4+x^2c^2$. I don't know what to do. – Aug 03 '20 at 08:20
@Navid Simplify it to $x^{2}(a^2-c^2)+y^{2}(a^2)=a^2(a^2-c^2)$ by rearranging terms. Then divide both sides by $a^2(a^2-c^2)$ . Do you get it? – Hrishabh Nayal Aug 03 '20 at 08:22
@HrishabhNayal Yes. Done! Thanks. – Aug 03 '20 at 08:28
@YvesDaoust I have deleted the downvote and I have given to the user a positive score. – Sebastiano Aug 03 '20 at 15:33

Sebastiano · Answer 3 · 2020-08-03T15:32:07.340

Starting of your definition in the question if $F_1(-c,0)$ and $F_2(c,0)$ are the foci ($c>0$) and $P\in\mathcal E$ with $P(x,y)$ and $O=(0,0)$ you have

$$\mathcal E: \quad |PF_1|+|PF_2|=2a$$

Hence:

$$(x+c)^2+y^2+(x-c)^2+y^2+2\sqrt{[(x+c)^2+y^2][(x-c)^2+y^2]}=4a^2$$

$$2x^2+2c^2+2y^2+2\sqrt{[(x+c)^2+y^2][(x-c)^2+y^2]} = 4a^2$$

$$x^2+c^2+y^2+\sqrt{[(x+c)^2+y^2][(x-c)^2+y^2]} = 2a^2$$

Isolate the root and square it again....

$$\sqrt{[(x+c)^2+y^2][(x-c)^2+y^2]} = 2a^2-x^2-y^2-c^2$$

$$(x^2+c^2+2xc+y^2) (x^2+c^2-2xc+y^2) = 4a^4+x^4+y^4+c^4-4a^2x^2-4a^2y^2-4a^2c^2+2x^2y^2+2x^2c^2+2y^2c^2$$

$$x^4+c^4+y^4+2x^2c^2+2x^2y^2+2c^2y^2-4x^2c^2 =4a^4+x^4+y^4+c^4-4a^2x^2-4a^2y^2-4a^2c^2+2x^2y^2+2x^2c^2+2y^2c^2$$

From which, simplifying,

$$4a^4-4a^2x^2-4a^2y^2-4a^2c^2+4x^2c^2 = 0$$

$$a^4-a^2x^2-a^2y^2-a^2c^2+x^2c^2 = 0$$

$$x^2(c^2-a^2)-a^2y^2 = a^2c^2 -a^4$$

Finally, remembering that $a^2-c^2 \gt 0$ we can put $b^2=a^2-c^2$ and have like this

$$-b^2x^2-a^2y^2=-b^2a^2$$ hence the proof

$$\begin{equation} \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \end{equation}$$

@YvesDaoust Surely someone who doesn't know the math ahahah has voted negative. — Sebastiano, Aug 03 '20 at 15:21

Proving the equation of an ellipse

3 Answers3