Given, We have to write an equation of an ellipse in standard form with the center at the origin
Given that vertex at (-3,0) and co-vertex (0,2)
The standard form of an ellipse is
where a is x- intercept and b is y – intercept.
We have vertex (-3, 0) and (0, 2) from these we can say that, x – intercept is – 3 and y – intercept is 2
. As we know that intercepts are the respective values when other variables becomes 0.
Given, We have to write an equation of an ellipse in standard form with the center at the origin
Given that vertex at (-3,0) and co-vertex (0,2)
The standard form of an ellipse is
where a is x- intercept and b is y – intercept.
We have vertex (-3, 0) and (0, 2) from these we can say that, x – intercept is – 3 and y – intercept is 2
. As we know that intercepts are the respective values when other variables becomes 0.
u basically just fill in the blank with a number that would add up to the number u already have that would equal the same as the other side. also pls give brainliest haha
u basically just fill in the blank with a number that would add up to the number u already have that would equal the same as the other side. also pls give brainliest haha
with the reciprocal of 
.
= 4/9
0
3 137
rating
![\bf \textit{ellipse, horizontal major axis} \\\\ \cfrac{(x- h)^2}{ a^2}+\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2- b ^2} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} h=0\\ k=0\\ a=3\\ b=2 \end{cases}\implies \cfrac{(x- 0)^2}{ 3^2}+\cfrac{(y- 0)^2}{ 2^2}=1\implies \cfrac{x^2}{9}+\cfrac{y^2}{4}=1](/tpl/images/0442/9503/82c42.png)
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