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!)
since the foci are simetrical the center of the ellipse is the middle point between the foci
in this case is quite easy. the center is (0,0)
so the equation u r looking for is like this \[ \large \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \]
b is the distance fro the center to either of the co-vertices can u find b?
i honestly do not know how to do any of this
ok just for future references. if u r given two point \(P(a,b)\) and \(Q(c,d)\) then the distance between these is \[ \large d(P,Q)=\sqrt{(a-c)^2+(b-d)^2} \]
beforre going on i have a question for u: how come u have to do excercises on something u've never been taught?
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
oh interesting i was just curious. sorry. did u understand the formula i posted
then your center is C(0,0) and one of the co-vertice is W1(0,3) then \[ \large b=d(C,W_1)=\sqrt{(0-0^2+(0-3)^2}=3 \]
so far \[ \large \frac{x^2}{a^2}+\frac{y^2}{3^2}=1 \]
yeah it's okay. I understood the formula I'm just trying to practice it now
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
Hello @SimplyLynn ???
sorry my laptop froze :/
lets finish this \[ \large c=\sqrt{(0-10)^2+(0-0)^2}=10 \] then \[ \large a^2-c^2=b^2 \] becomes \[ \large a^2-10^2=3^2 \] \[ \large a^2=9+100=109 \]
so finally \[ \large \frac{x^2}{109}+\frac{y^2}{9}=1 \]
so when we don't have the vertices given we use \[a ^{2}-c ^{2}=b ^{2}\] ?
yes when u don't have either the vertices, or the co-vertices, or the foci, u use that equation
okay, got it, thank you for your help and time :)
u r welcome good luck with your on-line course
thanks :)
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