Focus of an Ellipse

[HCL]

What is a focus of an ellipse with equation 9x2 + 25y2 = 900?
A. (0, 8)
B. (0, 6)
C. (6, 0)
D. (8, 0)

Answer is D but I'm not getting why.

 

[tkhunny]

Divide by 900.

 

[HallsofIvy]

If your equation were $\displaystyle \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$ would you be able to find the focus? If not you need to consult a textbook.

 

[HCL]

If your equation were $\displaystyle \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$ would you be able to find the focus? If not you need to consult a textbook.

i honestly don't have a textbook, i'm studying for a test but haven't taken math in a while... i'm mostly googling and answering questions but when i get stuck, i post. to the question tho, is the focus be (0,1)?? because a^2-b^2=c^2, c being the focus

 

[DrPhil]

i honestly don't have a textbook, i'm studying for a test but haven't taken math in a while... i'm mostly googling and answering questions but when i get stuck, i post. to the question tho, is the focus be (0,1)?? because a^2-b^2=c^2, c being the focus

Yes, that formula is something you have to do to find the focus. But first you have to find a and b.

$\displaystyle \displaystyle 9\ x^2 + 25\ y^2 = 900 \longrightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

You were given the clue to divide by 900. That looks like a good step toward finding a and b.

Next, you have to recognize which is larger, a or b. If a, then the focus is on the x-axis, and c^2 = a^2 - b^2. BUT if b is larger, then the focus is in the y-direction and c^2 = b^2 - a^2.

ok now?

 

[HCL]

lightbulb finally went off for part of this, i was wondering about dividing by 900 but i couldn't see why until you connected the x2/a+y2/b with the equation, I get it now, it's like the /a2 are already in the equation and you have to use the 900 to find a and b...

so i'm getting x2/9 + y2/36=1 so a=3 and b=6 but when i place c2=6^2-3^2=27... so the square root of 27 can be simplified to 3*square root3 but I'm still not seeing how to make the connection to (8,0). I understand it that C is the coordinate whether on y or x axis, in this case the y axis b/c b>a, but my c from the above is square root of 27... i feel like i have the right a and b, but maybe i'm messing up the formula?

Yes, that formula is something you have to do to find the focus. But first you have to find a and b.

$\displaystyle \displaystyle 100\ x^2 + 25\ y^2 = 900 \longrightarrow \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

You were given the clue to divide by 900. That looks like a good step toward finding a and b.

Next, you have to recognize which is larger, a or b. If a, then the focus is on the x-axis, and c^2 = a^2 - b^2. BUT if b is larger, then the focus is in the y-direction and c^2 = b^2 - a^2.

ok now?

 

[Deleted member 4993]

lightbulb finally went off for part of this, i was wondering about dividing by 900 but i couldn't see why until you connected the x2/a+y2/b with the equation, I get it now, it's like the /a2 are already in the equation and you have to use the 900 to find a and b...

so i'm getting x2/9 + y2/36=1 so a=3 and b=6 but when i place c2=6^2-3^2=27... so the square root of 27 can be simplified to 3*square root3 but I'm still not seeing how to make the connection to (8,0). I understand it that C is the coordinate whether on y or x axis, in this case the y axis b/c b>a, but my c from the above is square root of 27... i feel like i have the right a and b, but maybe i'm messing up the formula?

No... you need to be more careful !!

You started with

9x² + 25y² = 900

x²/10² + y²/6² = 1

Now continue.....

 

[HCL]

oh, i see... so then a = 10 and b = 6, so c^2=10^2-6^2, c=8, so it's (8,0)

i was making 10=a^2 instead of 10=a

these little mistakes are always my downfall. thanks for the help!

No... you need to be more careful !!

You started with

9x² + 25y² = 900

x²/10² + y²/6² = 1

Now continue.....

 

[HallsofIvy]

The foci of an ellipse have the property that the sum of the two distances from any point on the ellipse to the foci is a constant.

Suppose the ellipse $\displaystyle \frac{x^2}{a^2}+ \frac{y^2}{b^2}= 1$. Then the vertices are at (-a, 0), (a, 0), (0, -b), and (0, b). Assume that a> b so that the foci lie on the x-axis, between (-a, 0) and (a, 0). Call the foci (c, 0) and (-c, 0).

Look at the distances from the point (a, 0) to the foci. The distance from (a, 0) to (c, 0) is a- c. The distance from (a, 0) to (-c, 0) is a+ c. The sum of those two distances is 2a, of course.

Now look at the distances from the point (0, b) to the foci. The distance from (0, b) to (c, 0) is $\displaystyle \sqrt{b^2+ c^2}$. The distance from (0, -b) to (c, 0) is $\displaystyle \sqrt{b^2+ c^2}$. The sum of those two distances is $\displaystyle 2\sqrt{b^2+ c^2}$.

Since those two sums must be the same, we must have $\displaystyle 2\sqrt{b^2+ c^2}= 2a$. Cancelling the "2"s and squaring both sides, $\displaystyle b^2+ c^2= a^2$ or $\displaystyle c^2= a^2- b^2$.