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

Step2: Find the vertices and co - vertices

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

• The vertices (end - points of the major axis) are at \((0,\pm a)=(0,\pm3)\).
• The co - vertices (end - points of the minor axis) are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points and draw the ellipse

1. Plot the center \((0,0)\).
2. Plot the vertices \((0,3)\) and \((0, - 3)\).
3. Plot the co - vertices \((2,0)\) and \((-2,0)\).
4. Draw a smooth curve connecting these points to form the ellipse. The ellipse will be taller along the \(y\) - axis (since \(a = 3\) and \(b = 2\)) and symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

The graph is an ellipse centered at the origin with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\), and the ellipse is drawn by connecting these points smoothly. (To actually draw it, plot the center \((0,0)\), vertices \((0,3)\), \((0, - 3)\), co - vertices \((2,0)\), \((-2,0)\) and sketch the ellipse through them.)