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 \(a>b>0\) 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 \(y\)-intercepts (vertices, since major axis is along \(y\)-axis), set \(x = 0\). Then \(\frac{0^{2}}{4}+\frac{y^{2}}{9}=1\), so \(y^{2}=9\) and \(y=\pm3\). So the points are \((0, 3)\) and \((0,- 3)\).
• For the \(x\)-intercepts (co - vertices), set \(y = 0\). Then \(\frac{x^{2}}{4}+\frac{0^{2}}{9}=1\), 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 curve connecting these points to form the ellipse. The center of the ellipse is at the origin \((0,0)\) since there are no shifts in the \(x\) or \(y\) terms (the equation is of the form \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) with \(h = 0\) and \(k = 0\)).

Answer:

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

1. Plot the center at \((0,0)\).
2. Plot the \(y\)-intercepts (vertices) at \((0,3)\) and \((0, - 3)\).
3. Plot the \(x\)-intercepts (co - vertices) at \((2,0)\) and \((-2,0)\).
4. Draw a smooth, closed curve (ellipse) passing through these four points. The major axis is along the \(y\)-axis with length \(2a=6\) and the minor axis is along the \(x\)-axis with length \(2b = 4\).