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 \(a>b>0\) and the major axis is along the \(y\)-axis), where \(a^{2}=9\) and \(b^{2}=4\). So, \(a = 3\) and \(b=2\).

Step2: Find the vertices and co - vertices

• For the major axis (along \(y\)-axis): The vertices are at \((0,\pm a)=(0,\pm3)\).
• For the minor axis (along \(x\)-axis): The co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot the points \((0, 3)\), \((0,- 3)\), \((2,0)\) and \((-2,0)\) on the coordinate plane.
• Then, draw a smooth ellipse passing through these four points. The ellipse will be symmetric about both the \(x\)-axis and \(y\)-axis.

(Note: Since the problem asks to graph the equation, the final answer is the graph of the ellipse with vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\) as described above. If we were to describe the key points for the graph: vertices at \((0, 3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\) and the ellipse passing through them.)

Answer:

The graph is an ellipse with vertices at \((0, 3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\), symmetric about the \(x\) - axis and \(y\) - axis, passing through these four points. (To draw it, plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) and sketch a smooth ellipse through them.)