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Mathematics 7 Online
OpenStudy (anonymous):

What is the equation of the ellipse with foci (-10, 0), (10, 0) and co-vertices (0, -3), (0, 3). (I was never taught this, please help!)

OpenStudy (helder_edwin):

since the foci are simetrical the center of the ellipse is the middle point between the foci

OpenStudy (helder_edwin):

in this case is quite easy. the center is (0,0)

OpenStudy (helder_edwin):

so the equation u r looking for is like this \[ \large \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]

OpenStudy (helder_edwin):

b is the distance fro the center to either of the co-vertices can u find b?

OpenStudy (anonymous):

i honestly do not know how to do any of this

OpenStudy (helder_edwin):

ok just for future references. if u r given two point \(P(a,b)\) and \(Q(c,d)\) then the distance between these is \[ \large d(P,Q)=\sqrt{(a-c)^2+(b-d)^2} \]

OpenStudy (helder_edwin):

beforre going on i have a question for u: how come u have to do excercises on something u've never been taught?

OpenStudy (anonymous):

It's an online program I have to do for credit recovery so Conic Sections was thrown in there and I was never taught that unit in my regular high school class

OpenStudy (helder_edwin):

oh interesting i was just curious. sorry. did u understand the formula i posted

OpenStudy (helder_edwin):

then your center is C(0,0) and one of the co-vertice is W1(0,3) then \[ \large b=d(C,W_1)=\sqrt{(0-0^2+(0-3)^2}=3 \]

OpenStudy (helder_edwin):

so far \[ \large \frac{x^2}{a^2}+\frac{y^2}{3^2}=1 \]

OpenStudy (anonymous):

yeah it's okay. I understood the formula I'm just trying to practice it now

OpenStudy (helder_edwin):

now \(a\) is the distance from the center to either of the vertices but we don't have the vertices so we have to use the following relation \[ \large a^2-c^2=b^2 \] where \(c\) is the distance between the center and either of the foci. find this distance

OpenStudy (helder_edwin):

Hello @SimplyLynn ???

OpenStudy (anonymous):

sorry my laptop froze :/

OpenStudy (helder_edwin):

lets finish this \[ \large c=\sqrt{(0-10)^2+(0-0)^2}=10 \] then \[ \large a^2-c^2=b^2 \] becomes \[ \large a^2-10^2=3^2 \] \[ \large a^2=9+100=109 \]

OpenStudy (helder_edwin):

so finally \[ \large \frac{x^2}{109}+\frac{y^2}{9}=1 \]

OpenStudy (anonymous):

so when we don't have the vertices given we use \[a ^{2}-c ^{2}=b ^{2}\] ?

OpenStudy (helder_edwin):

yes when u don't have either the vertices, or the co-vertices, or the foci, u use that equation

OpenStudy (anonymous):

okay, got it, thank you for your help and time :)

OpenStudy (helder_edwin):

u r welcome good luck with your on-line course

OpenStudy (anonymous):

thanks :)

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