Question

graph each equation.

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

coordinate plane grid with x from -8 to 8 and y from -8 to 8

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 \(a^{2}>b^{2}\) 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 points.

(Note: Since the problem asks to graph the equation, the final answer is the graph of the ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above. But in text form, we can describe the steps to graph it as above.)

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 \((-2,0)\). To draw it, plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) and sketch a smooth curve through them.