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 \(a>b\) for vertical major axis), where \(a^{2}=9\) and \(b^{2}=4\), so \(a = 3\) and \(b=2\).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis, since \(a\) is under \(y^{2}\)): The vertices are at \((0,\pm a)=(0,\pm3)\).
• For the \(x\) - axis (minor axis): The co - vertices are at \((\pm b,0)=(\pm2,0)\).

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 ellipse is centered at the origin \((0,0)\) because there are no \(h\) or \(k\) shifts in the standard form \(\frac{(x - h)^{2}}{b^{2}}+\frac{(y - k)^{2}}{a^{2}}=1\) (here \(h = 0\) and \(k = 0\)).

Answer:

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

1. Identify it as an ellipse with vertical major axis (since \(a^{2}=9\) and \(b^{2}=4\), \(a > b\)), centered at \((0,0)\).
2. Plot vertices \((0,\pm3)\) and co - vertices \((\pm2,0)\).
3. Draw a smooth ellipse through these points. The graph is an ellipse centered at the origin, with vertices at \((0, 3)\), \((0, - 3)\) and co - vertices at \((2,0)\), \((-2,0)\).