Question

graph each equation.

1. \\(\frac{x^2}{4} + \frac{y^2}{9} = 1\\)

Explanation:

Step1: Identify the form of the equation

The equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\) is in the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (\(a>b>0\)). Here, \(a^{2}=9\), so \(a = 3\), and \(b^{2}=4\), so \(b = 2\).

Step2: Find the vertices

For the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the vertices are \((0,\pm a)\). Substituting \(a = 3\), the vertices are \((0,3)\) and \((0, - 3)\).

Step3: Find the co - vertices

The co - vertices are \((\pm b,0)\). Substituting \(b = 2\), the co - vertices are \((2,0)\) and \((-2,0)\).

Step4: Sketch the ellipse

Plot the vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\). Then draw a smooth curve passing through these four points to form the ellipse.

Answer:

The ellipse has vertices at \((0,3)\) and \((0, - 3)\), co - vertices at \((2,0)\) and \((-2,0)\). Sketch the ellipse passing through these four points.