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 \(a > b\), so major axis is along y - axis).

Step2: Find the vertices and co - vertices

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

• 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

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 will be symmetric about both the x - axis and y - axis.

Answer:

The graph is an ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\), plotted and connected as described above. (To actually draw it, plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) and draw a smooth curve through them, symmetric about the x - axis and y - axis.)