Question

write the equation of the ellipse that has its center at the origin and a focus at (0,2); major axis equal to 8. question help: video ebook

Explanation:

Step1: Determine the form of the ellipse equation

Since the center is at the origin \((0,0)\) and the focus is at \((0,2)\), the major axis is along the \(y -\)axis. The standard form of the ellipse equation is \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\) (\(a>b>0\)), where \(c\) is the distance from the center to the focus, \(a\) is the semi - major axis length, and \(b\) is the semi - minor axis length.

Step2: Find the value of \(a\)

Given that the length of the major axis \(2a = 8\), so \(a=4\).

Step3: Find the value of \(c\)

The focus is at \((0,2)\), so \(c = 2\).

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

Substitute \(a = 4\) and \(c = 2\) into \(c^{2}=a^{2}-b^{2}\). We get \(2^{2}=4^{2}-b^{2}\), which simplifies to \(b^{2}=a^{2}-c^{2}\). Then \(b^{2}=16 - 4=12\).

Step5: Write the equation of the ellipse

Substitute \(a^{2}=16\) and \(b^{2}=12\) into the standard form \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\). The equation is \(\frac{x^{2}}{12}+\frac{y^{2}}{16}=1\).

Answer:

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