Question

question 12 - 1 point
what is the standard form equation of the ellipse that has vertices (-8,2) and (6,2) and co - vertices (-1,-1) and (-1,5)?
select the correct answer below:
\\(\frac{x^{2}}{3}+\frac{y^{2}}{49}=1\\)
\\(\frac{(x + 1)^{2}}{49}+\frac{(y - 2)^{2}}{3}=1\\)
\\(\frac{(x + 1)^{2}}{49}+\frac{(y - 2)^{2}}{9}=1\\)
\\(\frac{(x + 1)^{2}}{9}+\frac{(y - 2)^{2}}{49}=1\\)
\\(\frac{(x + 1)^{2}}{3}+\frac{(y - 2)^{2}}{7}=1\\)
\\(\frac{(x + 1)^{2}}{7}+\frac{(y - 2)^{2}}{3}=1\\)

Explanation:

Step1: Find the center of the ellipse

The center of the ellipse \((h,k)\) is the mid - point of the vertices or co - vertices. Mid - point of vertices \((-8,2)\) and \((6,2)\) is \((\frac{-8 + 6}{2},\frac{2+2}{2})=(-1,2)\), so \(h=-1,k = 2\).

Step2: Find the value of \(a\)

The distance between the vertices \((-8,2)\) and \((6,2)\) is \(2a\). Using the distance formula \(d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\), here \(y_1=y_2 = 2\), so \(2a=\vert6-(-8)\vert=14\), then \(a = 7\), and \(a^2=49\).

Step3: Find the value of \(b\)

The distance between the co - vertices \((-1,-1)\) and \((-1,5)\) is \(2b\). Using the distance formula, here \(x_1=x_2=-1\), so \(2b=\vert5-(-1)\vert = 6\), then \(b = 3\), and \(b^2=9\).

Step4: Write the standard form of the ellipse equation

The standard form of an ellipse with a horizontal major axis is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\). Substituting \(h=-1,k = 2,a^2 = 49,b^2=9\) we get \(\frac{(x + 1)^2}{49}+\frac{(y - 2)^2}{9}=1\).

Answer:

\(\frac{(x + 1)^2}{49}+\frac{(y - 2)^2}{9}=1\)