Question

find an equation for the conic that satisfies the given conditions. ellipse, foci (±2, 0), vertices (±4, 0)

Explanation:

Step1: Determine the standard form of the ellipse equation

Since the foci and vertices are 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 vertices of the ellipse are \((\pm a,0)\). Given vertices \((\pm4,0)\), so \(a = 4\), and \(a^{2}=16\).

Step3: Find the value of \(c\)

The foci of the ellipse are \((\pm c,0)\). Given foci \((\pm2,0)\), so \(c = 2\).

Step4: Calculate the value of \(b^{2}\)

Use the relationship \(c^{2}=a^{2}-b^{2}\). Substitute \(a = 4\) and \(c = 2\) into it:

$$
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$$

Answer:

\(\frac{x^{2}}{16}+\frac{y^{2}}{12}=1\)