Question

what are the foci of the ellipse ( 5x^{2}+9y^{2}-45 = 0 )?
write your answer in simplified, rationalized form.
( (quad,quad) ) and ( (quad,quad) )

Explanation:

Step1: Write the ellipse equation in standard form

Divide the equation \(5x^{2}+9y^{2}-45 = 0\) by \(45\).
We get \(\frac{x^{2}}{9}+\frac{y^{2}}{5}=1\). For an ellipse of the form \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}} = 1\) (\(a>b>0\)), here \(a^{2}=9\), so \(a = 3\), and \(b^{2}=5\).

Step2: Calculate \(c\) using the formula \(c^{2}=a^{2}-b^{2}\)

Substitute \(a^{2}=9\) and \(b^{2}=5\) into \(c^{2}=a^{2}-b^{2}\).
We have \(c^{2}=9 - 5=4\), then \(c = 2\).

Step3: Determine the foci

Since the major axis is along the \(x\) - axis (because \(a^{2}\) is under \(x^{2}\)), the foci of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) are \((\pm c,0)\).

Answer:

\((- 2,0)\) and \((2,0)\)