Question

graph each equation.

1. \\(\dfrac{x^2}{4} + \dfrac{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 \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\) (where \(a > b>0\)) for an ellipse centered at the origin with a vertical major axis.

Step2: Find the values of \(a\) and \(b\)

For the given equation, we have \(a^{2}=9\) and \(b^{2} = 4\). Taking the square roots, we get \(a=\sqrt{9} = 3\) and \(b=\sqrt{4}=2\).

Step3: Determine the vertices and co - vertices

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

Step4: 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 will be symmetric about both the \(x\) - axis and \(y\) - axis.

Answer:

The graph is an ellipse centered at the origin with vertices at \((0, 3)\), \((0,-3)\) and co - vertices at \((2,0)\), \((- 2,0)\) (the actual drawing involves plotting these points and drawing a smooth elliptical curve through them).