Question

graph each equation.

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

Explanation:

Step1: Identify the conic section

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

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the center is at \((0,0)\) (the origin).

• The vertices are at \((0,\pm a)\). Since \(a^{2}=9\), then \(a = 3\), so the vertices are \((0,3)\) and \((0, - 3)\).
• The co - vertices are at \((\pm b,0)\). Since \(b^{2}=4\), then \(b = 2\), so the co - vertices are \((2,0)\) and \((- 2,0)\).

Step3: Plot the points and draw the ellipse

1. Plot the center \((0,0)\).
2. Plot the vertices \((0,3)\) and \((0,-3)\) (these are the top and bottom most points of the ellipse).
3. Plot the co - vertices \((2,0)\) and \((-2,0)\) (these are the right and left most points of the ellipse).
4. Then, draw a smooth curve connecting these points to form the ellipse. The ellipse will be taller along the \(y\) - axis (because \(a>b\)) and will pass through the points \((\pm2,0)\) and \((0,\pm3)\).

(Note: Since the question asks to graph the equation, the final answer is the graph of the ellipse with center at the origin, vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above. If we were to describe the key points for graphing: the ellipse is centered at \((0,0)\), has a vertical major axis with length \(2a = 6\) (from \(y=-3\) to \(y = 3\)) and a horizontal minor axis with length \(2b=4\) (from \(x=-2\) to \(x = 2\)).)

Answer:

The graph is an ellipse centered at the origin \((0,0)\), with vertices at \((0, 3)\) and \((0, - 3)\) and co - vertices at \((2, 0)\) and \((-2, 0)\). The ellipse is drawn by connecting these points with a smooth curve, being taller along the \(y\) - axis.