Question

graph each equation.

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

Explanation:

Step1: Identify the type of conic section

The equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\) is of the form \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\) (\(a>b>0\)), which represents an ellipse centered at the origin \((0,0)\).

Step2: Find the values of \(a\) and \(b\)

For the equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\), we have \(a^{2}=9\) and \(b^{2} = 4\). So, \(a = 3\) and \(b=2\).

Step3: Determine the vertices and co - vertices

The vertices of the ellipse (along the \(y\) - axis, since \(a^{2}\) is under \(y^{2}\)) are \((0,a)\) and \((0, - a)\), which are \((0,3)\) and \((0,-3)\). The co - vertices (along the \(x\) - axis) are \((b,0)\) and \((-b,0)\), which are \((2,0)\) and \((-2,0)\).

Step4: Sketch the ellipse

Plot the vertices \((0,3)\), \((0, - 3)\), co - vertices \((2,0)\) and \((-2,0)\). Then draw a smooth curve passing through these four points to form the ellipse.

Answer:

The graph is an ellipse centered at the origin with vertices \((0,3)\) and \((0,-3)\) and co - vertices \((2,0)\) and \((-2,0)\).