Question

graph each equation.

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

Explanation:

Step1: Identify the conic section type

The given equation is \(\frac{x^{2}}{4}+\frac{y^{2}}{9} = 1\), which is in the standard form of an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (since \(a^{2}=9\) and \(b^{2}=4\), so \(a = 3\), \(b = 2\)) with the major axis along the \(y\)-axis (because \(a>b\) and the \(y^{2}\) term has the larger denominator).

Step2: Find the vertices and co - vertices

• For the \(y\) - intercepts (vertices, since major axis is along \(y\) - axis): Set \(x = 0\) in the equation \(\frac{0^{2}}{4}+\frac{y^{2}}{9}=1\), then \(\frac{y^{2}}{9}=1\), so \(y^{2}=9\) and \(y=\pm3\). So the points are \((0, 3)\) and \((0, - 3)\).
• For the \(x\) - intercepts (co - vertices): Set \(y = 0\) in the equation \(\frac{x^{2}}{4}+\frac{0^{2}}{9}=1\), then \(\frac{x^{2}}{4}=1\), so \(x^{2}=4\) and \(x=\pm2\). So the points 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 symmetric about both the \(x\) - axis and \(y\) - axis.

To graph the ellipse:

1. Mark the points \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\) on the given coordinate grid.
2. Connect these points with a smooth curve, making sure the curve is symmetric with respect to both the \(x\) - axis and \(y\) - axis. The ellipse will be taller along the \(y\) - axis (from \(y=-3\) to \(y = 3\)) and narrower along the \(x\) - axis (from \(x=-2\) to \(x = 2\)).

Answer:

The graph is an ellipse with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\), plotted and connected as described above. (The actual graph is an ellipse centered at the origin, with major axis along the \(y\) - axis, passing through \((0,3)\), \((0, - 3)\), \((2,0)\) and \((-2,0)\))