Question

graph each equation.

1. \\(\dfrac{x^2}{4} + \dfrac{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 \(a>b\), major axis is along \(y\)-axis).

Step2: Find the vertices and co - vertices

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

• The vertices (end - points of the major axis) are \((0,\pm a)=(0, 3)\) and \((0,- 3)\).
• The co - vertices (end - points of the minor axis) are \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

Plot the points \((0,3)\), \((0, - 3)\), \((2,0)\) and \((- 2,0)\) on the coordinate plane.

Step4: Draw the ellipse

Connect the plotted points smoothly to form the ellipse. The ellipse is centered at the origin \((0,0)\) with the major axis along the \(y\)-axis (length \(2a = 6\)) and the minor axis along the \(x\)-axis (length \(2b=4\)).

(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 \((\pm2,0)\) centered at the origin.)

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((-2,0)\). To draw it, plot these four points and connect them smoothly to form the ellipse.