Question

graph each equation.

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

graph with x-axis from -8 to 8 and y-axis from -8 to 8, grid lines

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 \(a>b>0\)), so it is an ellipse centered at the origin \((0,0)\). Here, \(a^{2}=9\) so \(a = 3\) and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

For the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the vertices are at \((0,\pm a)\) and the co - vertices are at \((\pm b,0)\).

• Vertices: When \(x = 0\), from the equation \(\frac{0^{2}}{4}+\frac{y^{2}}{9}=1\), we get \(\frac{y^{2}}{9}=1\), so \(y^{2}=9\) and \(y=\pm3\). So the vertices are \((0, 3)\) and \((0,- 3)\).
• Co - vertices: When \(y = 0\), from the equation \(\frac{x^{2}}{4}+\frac{0^{2}}{9}=1\), we get \(\frac{x^{2}}{4}=1\), so \(x^{2}=4\) and \(x=\pm2\). So the co - vertices are \((2,0)\) and \((- 2,0)\).

Step3: Plot the points and draw the ellipse

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. The major axis is along the \(y\) - axis (since \(a\) is associated with the \(y\) - term) and the minor axis is along the \(x\) - axis.

Answer:

The graph is an ellipse centered at the origin with vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\). To draw it, plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) and sketch a smooth curve connecting them.