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 \(a>b\) 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 \(y\) - axis (major axis): When \(x = 0\), we solve for \(y\): \(\frac{0^{2}}{4}+\frac{y^{2}}{9}=1\Rightarrow y^{2}=9\Rightarrow y=\pm3\). So the vertices are \((0, 3)\) and \((0,- 3)\).
• For the \(x\) - axis (minor axis): When \(y = 0\), we solve for \(x\): \(\frac{x^{2}}{4}+\frac{0^{2}}{9}=1\Rightarrow x^{2}=4\Rightarrow x=\pm2\). 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 curve connecting these points to form the ellipse. The ellipse will be centered at the origin \((0,0)\) with a vertical major axis (since the denominator under \(y^{2}\) is larger) of length \(2a = 6\) and a horizontal minor axis of length \(2b=4\).

Answer:

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