Question

find the equation in standard form for each the ellipse with vertices at (±6,0) and foci at (±2,0).

Explanation:

Step1: Identify the value of a

The vertices of the ellipse are at $(\pm6,0)$. For an ellipse with a horizontal major - axis (since the vertices are on the x - axis), the distance from the center to a vertex is $a$. So, $a = 6$.

Step2: Identify the value of c

The foci of the ellipse are at $(\pm2,0)$. The distance from the center to a focus is $c$. So, $c = 2$.

Step3: Calculate the value of b

Use the relationship $c^{2}=a^{2}-b^{2}$ for an ellipse. Rearranging for $b^{2}$, we get $b^{2}=a^{2}-c^{2}$. Substitute $a = 6$ and $c = 2$ into the formula: $b^{2}=6^{2}-2^{2}=36 - 4=32$.

Step4: Write the standard - form equation of the ellipse

The standard - form equation of an ellipse with a horizontal major - axis is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1$. Substitute $a^{2}=36$ and $b^{2}=32$ into the equation: $\frac{x^{2}}{36}+\frac{y^{2}}{32}=1$.

Answer:

$\frac{x^{2}}{36}+\frac{y^{2}}{32}=1$