Question

find the standard form of the equation of the ellipse satisfying the given conditions. major axis horizontal with length 8; length of minor axis = 6; center: (0, 0) standard form of the equation: \boxed{}

Explanation:

Step1: Recall ellipse standard form

For an ellipse with center \((h,k)\), major axis horizontal, the standard form is \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\) where \(a>b\), \(2a\) is major axis length, \(2b\) is minor axis length. Here, center \((h,k)=(0,0)\), so \(h = 0\), \(k = 0\).

Step2: Find \(a\) and \(b\)

Major axis length \(2a = 8\), so \(a=\frac{8}{2}=4\), then \(a^2 = 16\). Minor axis length \(2b = 6\), so \(b=\frac{6}{2}=3\), then \(b^2 = 9\).

Step3: Substitute into standard form

Substitute \(h = 0\), \(k = 0\), \(a^2 = 16\), \(b^2 = 9\) into \(\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1\), we get \(\frac{x^2}{16}+\frac{y^2}{9}=1\).

Answer:

\(\frac{x^2}{16}+\frac{y^2}{9}=1\)