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 larger denominator is under \(y^{2}\)).

Step2: Find the vertices and co - vertices

• For the \(y\)-intercepts (vertices along the major axis): Set \(x = 0\), then \(\frac{y^{2}}{9}=1\), so \(y^{2}=9\), \(y=\pm3\). So the points are \((0, 3)\) and \((0,- 3)\).
• For the \(x\)-intercepts (co - vertices along the minor axis): Set \(y = 0\), then \(\frac{x^{2}}{4}=1\), so \(x^{2}=4\), \(x=\pm2\). 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 curve connecting these points to form the ellipse. The ellipse will be centered at the origin \((0,0)\) (since there are no \(h\) and \(k\) shifts in the standard form \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\), here \(h = 0\) and \(k = 0\)).

Answer:

To graph \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\):

1. Recognize it as an ellipse centered at \((0,0)\) with major axis along the \(y\)-axis, \(a = 3\), \(b = 2\).
2. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
3. Draw a smooth ellipse through these points. The graph is an ellipse centered at the origin, with \(y\)-intercepts at \((0,\pm3)\) and \(x\)-intercepts at \((\pm2,0)\).