Question

Write the equation of an ellipse with vertices (-5,1) and (-1,1) and co-vertices (-3,2) and (-3,0)

Please explain.

Answer 1

Answer: To find the equation of the ellipse, we need to use the standard form equation of an ellipse:

$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$

where (h,k) is the center of the ellipse, a is the distance from the center to the vertices (the major axis), and b is the distance from the center to the co-vertices (the minor axis).

First, let's find the center of the ellipse. The center of the ellipse is the midpoint of the line segment joining the vertices (-5,1) and (-1,1). Using the midpoint formula, we get:

$(h,k) = \left(\frac{-5+(-1)}{2}, 1\right) = (-3,1)$

Now, we need to find the values of a and b. Since the distance between the vertices is 2a, we have:

$2a = |-5-(-1)| = 4$

So, $a = 2$.

Similarly, the distance between the co-vertices is 2b, we have:

$2b = |2-0| = 2$

So, $b = 1$.

Now, we have all the values we need to write the equation of the ellipse:

$\frac{(x+3)^2}{2^2}+\frac{(y-1)^2}{1^2}=1$

Simplifying this equation, we get:

$\frac{(x+3)^2}{4}+(y-1)^2=1$

So, the equation of the ellipse is:

$(x+3)^2/4 + (y-1)^2/1 = 1$

Explanation:

Answer 2

Check the picture below.

`\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) \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \begin{cases} h=-3\\ k=1\\ a=2\\ b=1 \end{cases}\implies \cfrac{(x- (-3))^2}{ 2^2}+\cfrac{(y-1)^2}{ 1^2}=1\implies \cfrac{(x+3)^2}{ 4}+\cfrac{(y-1)^2}{ 1}=1`