Question

what are the co - vertices of the ellipse (5x^{2}+2y^{2}-20 = 0)?
write your answer in simplified, rationalized form.
((quad,quad)) and ((quad,quad))

Explanation:

Step1: Rewrite the ellipse equation in standard form

Given \(5x^{2}+2y^{2}-20 = 0\), add \(20\) to both sides: \(5x^{2}+2y^{2}=20\). Divide by \(20\): \(\frac{x^{2}}{4}+\frac{y^{2}}{10}=1\). For an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\) (\(a>b>0\)), here \(a^{2}=10\), \(b^{2}=4\), so \(a=\sqrt{10}\), \(b = 2\).

Step2: Determine the co - vertices

For the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the co - vertices are \((\pm b,0)\). Since \(b = 2\), the co - vertices are \((- 2,0)\) and \((2,0)\).

Answer:

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