Question

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

Explanation:

Step1: Identify the conic section

The 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 \(a>b\) and major axis is along y - axis). Here, \(a^{2}=9\) so \(a = 3\) and \(b^{2}=4\) so \(b = 2\).

Step2: Find the vertices and co - vertices

• For an ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the vertices (end - points of the major axis) are \((0,\pm a)=(0,\pm3)\) and the co - vertices (end - points of the minor axis) are \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

• Plot 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 centered at the origin \((0,0)\) since there are no shifts in the \(x\) or \(y\) terms (the equation is in the form \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) with \(h = 0\) and \(k = 0\)).

To graph the ellipse:

1. Mark the center at \((0,0)\).
2. Mark the co - vertices at \((2,0)\) and \((-2,0)\) (on the x - axis, 2 units from the center).
3. Mark the vertices at \((0,3)\) and \((0,-3)\) (on the y - axis, 3 units from the center).
4. Draw a smooth elliptical curve passing through these four points.

Answer:

The graph is an ellipse centered at the origin with vertices at \((0,\pm3)\) and co - vertices at \((\pm2,0)\) (the actual graph is a smooth curve connecting these points as described above).