Question

graph each equation.

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

Explanation:

Step1: Identify the ellipse type

The equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\), which is in the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (since \(a^{2}=9\) and \(b^{2}=4\), so \(a = 3\), \(b = 2\)). The major axis is vertical (because \(a^{2}\) is under \(y^{2}\)).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis) vertices: When \(x = 0\), we solve for \(y\) from \(\frac{0^{2}}{4}+\frac{y^{2}}{9}=1\), so \(y^{2}=9\), \(y=\pm3\). So the vertices are \((0, 3)\) and \((0, - 3)\).
• For the \(x\) - axis (minor axis) co - vertices: When \(y = 0\), we solve for \(x\) from \(\frac{x^{2}}{4}+\frac{0^{2}}{9}=1\), so \(x^{2}=4\), \(x=\pm2\). So the co - vertices are \((2, 0)\) and \((- 2, 0)\).

Step3: Plot the points and draw the ellipse

Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane. Then, sketch the ellipse passing through these points, with the major axis along the \(y\) - axis and minor axis along the \(x\) - axis.

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0, 3)\), \((0, - 3)\) and co - vertices at \((2, 0)\), \((-2, 0)\). To draw it, plot these four points and sketch the ellipse connecting them, with the major axis vertical (length \(2a = 6\)) and minor axis horizontal (length \(2b=4\)).