Question

what are the vertices of the ellipse ( 25x^{2}+3y^{2}-75 = 0 )?
write your answer in simplified, rationalized form.
( , ) and ( , )

Explanation:

Step1: Rewrite the equation in standard form

The standard form of an ellipse is \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\).
Starting with \(25x^{2}+3y^{2}-75 = 0\), we can rewrite it as \(25x^{2}+3y^{2}=75\).
Divide each term by \(75\): \(\frac{25x^{2}}{75}+\frac{3y^{2}}{75}=1\), which simplifies to \(\frac{x^{2}}{3}+\frac{y^{2}}{25}=1\).

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

For the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\), here \(a^{2}=3\), \(b^{2}=25\), so \(a=\sqrt{3}\), \(b = 5\).
Since \(b>a\), the major axis is along the \(y\) - axis. The vertices of an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) (\(b > a\)) are \((0,\pm b)\).

Answer:

\((0,5)\) and \((0, - 5)\)