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

Question

graph each equation.

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

coordinate grid image omitted

Explanation:

Step1: Identify the ellipse standard form

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

Step2: Find the vertices and co - vertices

For the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), where \(a=\sqrt{9} = 3\) and \(b=\sqrt{4}=2\).

  • The vertices (end - points of the major axis) are at \((0,\pm a)=(0,\pm3)\).
  • The co - vertices (end - points of the minor axis) are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

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.

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

  1. Recognize it is an ellipse with major axis along the \(y\) - axis (since the denominator of \(y^{2}\) is larger).
  2. Determine \(a = 3\) (distance from center \((0,0)\) to vertices on \(y\) - axis) and \(b=2\) (distance from center to co - vertices on \(x\) - axis).
  3. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
  4. Draw a smooth elliptical curve through these four points.

The graph will be an ellipse centered at the origin \((0,0)\), with the top and bottom points at \((0,3)\) and \((0, - 3)\) and the left and right points at \((-2,0)\) and \((2,0)\), and the curve connecting them smoothly.

Answer:

Step1: Identify the ellipse standard form

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

Step2: Find the vertices and co - vertices

For the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), where \(a=\sqrt{9} = 3\) and \(b=\sqrt{4}=2\).

  • The vertices (end - points of the major axis) are at \((0,\pm a)=(0,\pm3)\).
  • The co - vertices (end - points of the minor axis) are at \((\pm b,0)=(\pm2,0)\).

Step3: Plot the points

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.

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

  1. Recognize it is an ellipse with major axis along the \(y\) - axis (since the denominator of \(y^{2}\) is larger).
  2. Determine \(a = 3\) (distance from center \((0,0)\) to vertices on \(y\) - axis) and \(b=2\) (distance from center to co - vertices on \(x\) - axis).
  3. Plot the vertices \((0,3)\), \((0, - 3)\) and co - vertices \((2,0)\), \((-2,0)\).
  4. Draw a smooth elliptical curve through these four points.

The graph will be an ellipse centered at the origin \((0,0)\), with the top and bottom points at \((0,3)\) and \((0, - 3)\) and the left and right points at \((-2,0)\) and \((2,0)\), and the curve connecting them smoothly.