Question

graph each equation. 9) \\(\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 \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (where \(a > b>0\)) for an ellipse centered at the origin with a vertical major 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 \(y\). From \(\frac{0^{2}}{4}+\frac{y^{2}}{9}=1\), we get \(y^{2}=9\), so \(y=\pm3\). So the vertices are \((0, 3)\) and \((0,- 3)\).
• For the \(x\) - axis (minor axis) co - vertices: When \(y = 0\), we solve for \(x\). From \(\frac{x^{2}}{4}+\frac{0^{2}}{9}=1\), we get \(x^{2}=4\), so \(x=\pm2\). 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 taller along the \(y\) - axis (since \(a = 3\) and \(b = 2\)) and symmetric about both the \(x\) - axis and \(y\) - axis.

(Note: Since the question is about graphing, the final answer is the graph of the ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above. If we were to describe the key points for graphing: )

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 sketch a smooth, symmetric curve (ellipse) passing through them.