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 \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) (since \(a^2 = 9\) and \(b^2=4\), and \(a>b\), so it is a vertical ellipse).

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\), the center is at \((0,0)\) (the origin).

• The length of the semi - major axis \(a=\sqrt{9} = 3\), so the vertices are at \((0,\pm a)=(0, 3)\) and \((0,- 3)\)? Wait, no, \(a^2 = 9\) so \(a = 3\), and the vertices are \((0,\pm a)=(0,3)\) and \((0, - 3)\)? Wait, no, wait: the standard form for a vertical ellipse (major axis along y - axis) is \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\), where \((h,k)\) is the center. Here \(h = 0,k = 0\), \(a=\sqrt{9}=3\), \(b=\sqrt{4} = 2\). So the vertices (endpoints of the major axis) are \((0,k\pm a)=(0,0\pm3)=(0,3)\) and \((0, - 3)\)? Wait, no, that's wrong. Wait, if the major axis is along the y - axis, then the vertices are \((0,\pm a)\) and the co - vertices are \((\pm b,0)\). So \(a = 3\), so vertices are \((0,3)\) and \((0, - 3)\)? Wait, no, \(a^2=9\), so \(a = 3\), so the distance from the center to the vertices along the y - axis is \(a = 3\), so vertices are \((0,3)\) and \((0, - 3)\)? Wait, no, that can't be. Wait, no, the standard equation of an ellipse with center \((h,k)\), major axis length \(2a\) (along y - axis) and minor axis length \(2b\) (along x - axis) is \(\frac{(x - h)^2}{b^2}+\frac{(y - k)^2}{a^2}=1\). So when \(x = 0\), \(y=\pm a\), and when \(y = 0\), \(x=\pm b\). So when \(y = 0\), \(\frac{x^2}{4}=1\), so \(x^2=4\), \(x=\pm2\). When \(x = 0\), \(\frac{y^2}{9}=1\), so \(y^2 = 9\), \(y=\pm3\). So the co - vertices (endpoints of the minor axis) are \((\pm2,0)\) and the vertices (endpoints of the major axis) are \((0,\pm3)\).

Step3: Plot the points

• Plot the center \((0,0)\).
• Plot the co - vertices \((2,0)\) and \((- 2,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\).
• Then draw an 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 center at \((0,0)\).
2. Mark the points \((2,0)\), \((- 2,0)\), \((0,3)\) and \((0, - 3)\).
3. Sketch a smooth curve connecting these points, making sure the ellipse is wider along the y - axis (since \(a = 3\) and \(b = 2\), the major axis is vertical) and symmetric with respect to both the x - axis and y - axis.

(Note: Since the problem asks to graph the equation, the final answer is the graph of the ellipse with center at the origin, vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) as described above.)

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((- 2,0)\). The ellipse is drawn by connecting these points with a smooth curve, symmetric about both the \(x\) - axis and \(y\) - axis.