Question

graph each equation.

1. $\frac{x^2}{4} + \frac{y^2}{9} = 1$

Explanation:

Step1: Identify the conic section type

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 \(a>b>0\) and the major axis is along the \(y\)-axis). Here, \(a^{2}=9\) so \(a = 3\), and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis) vertices: When \(x = 0\), we solve for \(y\): \(\frac{0^{2}}{4}+\frac{y^{2}}{9}=1\), so \(y^{2}=9\) and \(y=\pm3\). So the vertices are \((0, 3)\) and \((0,- 3)\).
• For the \(x\) - axis (minor axis) co - vertices: When \(y = 0\), we solve for \(x\): \(\frac{x^{2}}{4}+\frac{0^{2}}{9}=1\), 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 ellipse will be taller along the \(y\) - axis (since \(a = 3\) and \(b = 2\)) with the center at the origin \((0,0)\).

Answer:

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