Question

graph each equation.

1. \\(\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 \(a>b>0\)), which is a vertical ellipse centered at the origin \((0,0)\).

Step2: Find the values of \(a\) and \(b\)

For the equation \(\frac{x^{2}}{4}+\frac{y^{2}}{9}=1\), we have \(b^{2} = 4\) so \(b=\sqrt{4} = 2\), and \(a^{2}=9\) so \(a=\sqrt{9}=3\).

Step3: Determine the vertices and co - vertices

• The vertices of a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) are at \((0,\pm a)\). So the vertices are \((0, 3)\) and \((0,- 3)\).
• The co - vertices are at \((\pm b,0)\). So the co - vertices are \((2,0)\) and \((- 2,0)\).

Step4: Plot the points and draw the ellipse

Plot the center \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((-2,0)\). Then draw a smooth curve connecting these points to form the ellipse. The ellipse will be taller along the \(y\) - axis (since \(a = 3\) is associated with the \(y\) - term) and wider along the \(x\) - axis with a width of \(2b=4\) (from \(x=-2\) to \(x = 2\)) and a height of \(2a = 6\) (from \(y=-3\) to \(y = 3\)).

Answer:

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

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