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 \(b^{2}=4\), so \(a = 3\), \(b = 2\)) with a vertical major axis (because \(a^{2}\) is under the \(y^{2}\) term).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis) vertices: When \(x = 0\), from \(\frac{0^{2}}{4}+\frac{y^{2}}{9}=1\), we get \(y^{2}=9\), so \(y=\pm3\). So the vertices are \((0, 3)\) and \((0,- 3)\).
• For the \(x\) - axis (minor axis) co - vertices: When \(y = 0\), from \(\frac{x^{2}}{4}+\frac{0^{2}}{9}=1\), we get \(x^{2}=4\), so \(x=\pm2\). 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, draw a smooth curve connecting these points to form the ellipse. The ellipse will be centered at the origin \((0,0)\) (since there are no \(x\) or \(y\) shifts in the standard form equation), with a vertical major axis of length \(2a=6\) and a horizontal minor axis of length \(2b = 4\).

To graph the ellipse:

1. Mark the center at \((0,0)\).
2. Mark the vertices at \((0,3)\) and \((0, - 3)\) (along the \(y\) - axis, 3 units up and down from the center).
3. Mark the co - vertices at \((2,0)\) and \((-2,0)\) (along the \(x\) - axis, 2 units left and right from the center).
4. Draw a smooth, closed curve passing through these four points, which is the graph of the ellipse \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\).

Answer:

The graph is an ellipse centered at the origin with vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\), and it is drawn by connecting these points with a smooth curve. (The actual graphing is done by plotting the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) and drawing the ellipse through them on the given coordinate grid.)