Question

write the standard form of an equation of an ellipse subject to the given conditions vertices: (4,0) and (-4,0); foci: (3,0) and (-3,0)

Explanation:

Step1: Determine the form of the ellipse equation

Since the vertices \((\pm4,0)\) and foci \((\pm3,0)\) lie on the \(x -\)axis, the standard form of the ellipse equation is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\) (\(a>b>0\)).

Step2: Find the value of \(a\)

The distance from the center to a vertex is \(a\). Given vertices \((\pm4,0)\), so \(a = 4\), and \(a^{2}=16\).

Step3: Find the value of \(c\)

The distance from the center to a focus is \(c\). Given foci \((\pm3,0)\), so \(c = 3\).

Step4: Use the relationship \(c^{2}=a^{2}-b^{2}\) to find \(b^{2}\)

Substitute \(a = 4\) and \(c = 3\) into \(c^{2}=a^{2}-b^{2}\). Then \(3^{2}=4^{2}-b^{2}\), which simplifies to \(b^{2}=a^{2}-c^{2}\). So \(b^{2}=16 - 9=7\).

Answer:

The standard form of the ellipse equation is \(\frac{x^{2}}{16}+\frac{y^{2}}{7}=1\)