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

Question

graph each equation.

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

graph with x-axis from -8 to 8 and y-axis from -8 to 8, grid lines, and origin at (0,0)

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 \((0,0)\) (since there are no shifts in \(x\) or \(y\) from the origin).

  • The length of the semi - major axis \(a=\sqrt{9}=3\), so the vertices are at \((0,\pm a)=(0, 3)\) and \((0,- 3)\).
  • The length of the semi - minor axis \(b = \sqrt{4}=2\), so the co - vertices are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

Plot the center \((0,0)\), the vertices \((0,3)\), \((0, - 3)\) and the co - vertices \((2,0)\), \((- 2,0)\). Then, sketch the ellipse passing through these points. The ellipse will be symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

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

  1. Recognize it is a vertical ellipse centered at the origin \((0,0)\) with \(a = 3\) (semi - major axis) and \(b=2\) (semi - minor axis).
  2. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
  3. Sketch the ellipse passing through these points, symmetric about the \(x\) - axis and \(y\) - axis. The graph is an ellipse centered at the origin, stretching 3 units up and down from the center, and 2 units left and right from the center.