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 \(b^2=4\), so \(a = 3\), \(b = 2\)) with a vertical major axis (because \(a>b\) and the \(y^2\) term has the larger denominator).

Step2: Find the vertices and co - vertices

• For the \(y\) - axis (major axis) vertices: When \(x = 0\), we solve for \(y\). Substituting \(x = 0\) into the equation \(\frac{0^2}{4}+\frac{y^2}{9}=1\), we get \(y^2=9\), so \(y=\pm3\). So the vertices are \((0, 3)\) and \((0,- 3)\).
• For the \(x\) - axis (minor axis) co - vertices: When \(y = 0\), we solve for \(x\). Substituting \(y = 0\) into the equation \(\frac{x^2}{4}+\frac{0^2}{9}=1\), we get \(x^2 = 4\), so \(x=\pm2\). So the co - vertices are \((2,0)\) and \((- 2,0)\).

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 ellipse passing through these points. The ellipse will be centered at the origin \((0,0)\), stretched 3 units up and down along the \(y\) - axis and 2 units left and right along the \(x\) - axis.

Answer:

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

1. Plot the vertices \((0, 3)\) and \((0,-3)\) (on the \(y\) - axis, 3 units from the origin).
2. Plot the co - vertices \((2,0)\) and \((-2,0)\) (on the \(x\) - axis, 2 units from the origin).
3. Draw a smooth ellipse passing through these four points, centered at the origin with a vertical major axis.