Question

graph each equation.

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

Explanation:

Step1: Identify the conic section type

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's a vertical ellipse centered at the origin \((0,0)\)).

Step2: Find the vertices and co - vertices

For a vertical ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\), the vertices are at \((0,\pm a)\) and the co - vertices are at \((\pm b,0)\).
Given \(a^{2}=9\), then \(a = 3\), so the vertices are \((0,3)\) and \((0, - 3)\)? Wait, no, \(a^{2}=9\) implies \(a = 3\)? Wait, no, \(a^{2}=9\) so \(a=\sqrt{9}=3\)? Wait, no, the standard form for vertical ellipse is \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) with center \((h,k)\). Here \(h = 0,k = 0\), \(a^{2}=9\) so \(a = 3\), \(b^{2}=4\) so \(b = 2\). So the vertices (end - points of the major axis) are \((0,\pm a)=(0,\pm3)\)? Wait, no, \(a\) is the semi - major axis length. Wait, \(a^{2}=9\), so \(a = 3\), so the vertices are \((0,3)\) and \((0, - 3)\)? No, wait, if \(a^{2}\) is under the \(y\) - term, the major axis is along the \(y\) - axis. So the vertices (the endpoints of the major axis) are \((0,a)\) and \((0, - a)\) where \(a=\sqrt{9}=3\), so \((0,3)\) and \((0, - 3)\)? Wait, no, \(a^{2}=9\) means \(a = 3\), so the top vertex is \((0,3)\) and bottom vertex is \((0, - 3)\)? Wait, no, \(a^{2}=9\) so \(a = 3\), so the distance from the center to the top and bottom of the ellipse is 3. The co - vertices (end - points of the minor axis) are \((b,0)\) and \((-b,0)\) where \(b = 2\), so \((2,0)\) and \((-2,0)\).

Step3: Plot the points

• Plot the center \((0,0)\).
• Plot the vertices \((0,3)\) and \((0, - 3)\) (wait, no, wait \(a^{2}=9\), so \(a = 3\), so the vertices are \((0,3)\) and \((0, - 3)\)? Wait, no, \(a^{2}=9\) implies \(a = 3\), so the major axis is along the \(y\) - axis with length \(2a=6\), from \(y=- 3\) to \(y = 3\)? Wait, no, \(a\) is the semi - major axis. So the top vertex is \((0,a)=(0,3)\) and the bottom vertex is \((0, - a)=(0, - 3)\)? Wait, no, \(a^{2}=9\), so \(a = 3\), so the vertices are \((0,3)\) and \((0, - 3)\), and the co - vertices are \((2,0)\) and \((-2,0)\).

Step4: Sketch the ellipse

Connect the points \((0,3)\), \((2,0)\), \((0, - 3)\), \((-2,0)\) with a smooth curve to form the ellipse.

Answer:

The graph is an ellipse centered at the origin \((0,0)\) with vertices at \((0,3)\) and \((0, - 3)\) and co - vertices at \((2,0)\) and \((-2,0)\), and the ellipse is drawn by connecting these points with a smooth curve. (To actually draw it on the given grid, plot the points \((0,3)\), \((0, - 3)\), \((2,0)\), \((-2,0)\) and then sketch the ellipse passing through these points.)