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?

Answer:

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

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

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- Feb 21st 2010, 02:40 PMMathBlaster47Equation 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?

Answer:

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

I think I have it right, but it never hurts to make sure... - Feb 21st 2010, 03:19 PMSoroban
Hello, MathBlaster47!

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

Quote:

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

- Feb 21st 2010, 03:26 PMMathBlaster47
- Feb 22nd 2010, 06:24 PMStonerPenguin
Um, actually there's one little error;

It's $\displaystyle \frac{x^2}{9}+\frac{y^2}{25}=1$ not $\displaystyle \frac{x^2}{25}+\frac{y^2}{9}=1$ 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),. You followed the guideline for an ellipse having center (0, 0),__foci (0, -c) and (0, c__).__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;*

**..............$\displaystyle \frac{x^2}{b^2}+\frac{y^2}{a^2}=1$**

**where $\displaystyle b^2 = a^2 - c^2$ "**

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 $\displaystyle b^2 = a^2 - c^2$ ;

$\displaystyle 5^2 - 4^2 = b^2$

$\displaystyle 25 - 16 = 9$ so b = 3

Since the foci are on the y-axis, the major axis is vertical, so use the appropriate formula;

$\displaystyle \frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ plug in the values;

$\displaystyle \frac{x^2}{3^2}+\frac{y^2}{5^2}=1$

$\displaystyle \frac{x^2}{9}+\frac{y^2}{25}=1$

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 (Rofl) but then I double checked :D Hoorah~ - Feb 23rd 2010, 05:36 AMMathBlaster47