# Thread: Equation of an ellipse

1. ## Equation of an ellipse

Is my answer to this question correct?

What is the equation of the ellipse with foci at (0,-4) and (0,4) ant the sum of its focal radii being 10?

\frac{x}{25}+\frac{y}{9}=1

I think I have it right, but it never hurts to make sure...

2. Hello, MathBlaster47!

Probably an oversight . . . you forgot the "squares".

Is my answer to this question correct?

What is the equation of the ellipse with foci at (0,-4) and (0,4)
and the sum of its focal radii being 10?

Answer: . $\frac{x^{\color{red}2}}{25}+\frac{y^{{\color{red}2 }}}{9}\:=\:1$

3. Originally Posted by Soroban
Hello, MathBlaster47!

Probably an oversight . . . you forgot the "squares".

Indeed....an oversight, along with the [tex] tags.....
Must remember to type slower!
But the bright side is that I'm correct!

4. Originally Posted by MathBlaster47
Indeed....an oversight, along with the [tex] tags.....
Must remember to type slower!
But the bright side is that I'm correct!
Um, actually there's one little error;
It's $\frac{x^2}{9}+\frac{y^2}{25}=1$ not $\frac{x^2}{25}+\frac{y^2}{9}=1$ because it's an ellipse streched along the y-axis. Look at page 420, they show two guidelines (in grey); the one on the left shows the guideline for an ellipse having center (0, 0), foci (0, -c) and (0, c). You followed the guideline for an ellipse having center (0, 0), foci (c, 0) and (c, 0).

I'll show ya;
"The ellipse having center (0, 0), foci (0, -c) and (0, c), and sum of focal radii 2a has the equation;
.............. $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
where $b^2 = a^2 - c^2$ "

So, to go back to the orginal question;
What is the equation of the ellipse with foci at (0,-4) and (0,4) ant the sum of its focal radii being 10?

So 10 = 2a, so a = 5 and c = 4. Now plug those values in $b^2 = a^2 - c^2$ ;
$5^2 - 4^2 = b^2$
$25 - 16 = 9$ so b = 3
Since the foci are on the y-axis, the major axis is vertical, so use the appropriate formula;
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ plug in the values;
$\frac{x^2}{3^2}+\frac{y^2}{5^2}=1$
$\frac{x^2}{9}+\frac{y^2}{25}=1$

Sorry if this is a little lengthy, I just wanted to make sure for the both of us. I actually erased my answers and wrote yours when I saw that ours were different but then I double checked Hoorah~

5. Originally Posted by StonerPenguin
Um, actually there's one little error;
It's $\frac{x^2}{9}+\frac{y^2}{25}=1$ not $\frac{x^2}{25}+\frac{y^2}{9}=1$ because it's an ellipse streched along the y-axis. Look at page 420, they show two guidelines (in grey); the one on the left shows the guideline for an ellipse having center (0, 0), foci (0, -c) and (0, c). You followed the guideline for an ellipse having center (0, 0), foci (c, 0) and (c, 0).

I'll show ya;
"The ellipse having center (0, 0), foci (0, -c) and (0, c), and sum of focal radii 2a has the equation;
.............. $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$
where $b^2 = a^2 - c^2$ "

So, to go back to the orginal question;
What is the equation of the ellipse with foci at (0,-4) and (0,4) ant the sum of its focal radii being 10?

So 10 = 2a, so a = 5 and c = 4. Now plug those values in $b^2 = a^2 - c^2$ ;
$5^2 - 4^2 = b^2$
$25 - 16 = 9$ so b = 3
Since the foci are on the y-axis, the major axis is vertical, so use the appropriate formula;
$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ plug in the values;
$\frac{x^2}{3^2}+\frac{y^2}{5^2}=1$
$\frac{x^2}{9}+\frac{y^2}{25}=1$

Sorry if this is a little lengthy, I just wanted to make sure for the both of us. I actually erased my answers and wrote yours when I saw that ours were different but then I double checked Hoorah~
No problem, better "a little lengthy" than too short, confusing and frustration inducing.
That said, many thanks for helping out!
As a lolcat might say....Helpin' yous doin it rite!