Question

find the vertices, co - vertices, foci, and eccentricity.
4.11)
\\(\frac{x^{2}}{49}+\frac{y^{2}}{9}=1\\)
vertices
co - vertices

Explanation:

Step1: Identify the ellipse standard form

The given equation is \(\frac{x^{2}}{49}+\frac{y^{2}}{9} = 1\), which is in the standard form of an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) where \(a > b>0\). So, \(a^{2}=49\) and \(b^{2}=9\), which means \(a = 7\) and \(b = 3\).

Step2: Find the vertices

For an ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) (major axis along the x - axis), the vertices are \((\pm a,0)\). Substituting \(a = 7\), the vertices are \((7,0)\) and \((- 7,0)\).

Step3: Find the co - vertices

For the same ellipse, the co - vertices are \((0,\pm b)\). Substituting \(b = 3\), the co - vertices are \((0,3)\) and \((0,-3)\).

Step4: Find the foci

First, we calculate \(c\) using the relationship \(c^{2}=a^{2}-b^{2}\). Substituting \(a^{2}=49\) and \(b^{2}=9\), we get \(c^{2}=49 - 9=40\), so \(c=\sqrt{40}=2\sqrt{10}\). The foci of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) (major axis along x - axis) are \((\pm c,0)\), so the foci are \((2\sqrt{10},0)\) and \((-2\sqrt{10},0)\).

Step5: Find the eccentricity

The formula for the eccentricity \(e\) of an ellipse is \(e=\frac{c}{a}\). We know that \(c = 2\sqrt{10}\) and \(a = 7\), so \(e=\frac{2\sqrt{10}}{7}\).

Answer:

• Vertices: \((7,0)\), \((-7,0)\)
• Co - vertices: \((0,3)\), \((0,-3)\)
• Foci: \((2\sqrt{10},0)\), \((-2\sqrt{10},0)\)
• Eccentricity: \(\frac{2\sqrt{10}}{7}\)