Graph (x^2)/9+(y^2)/16=1

Problem

(x^2)/9+(y^2)/16=1

Solution

1. Identify the type of conic section. Since the equation is in the form (x^2)/(a^2)+(y^2)/(b^2)=1 with positive denominators, it is an ellipse centered at the origin (0,0)

2. Determine the lengths of the semi-axes. The denominator under x^2 is a^2=9 so a=3 The denominator under y^2 is b^2=16 so b=4

3. Locate the vertices and co-vertices. Since b>a the major axis is vertical. The vertices are at (0,±4) and the co-vertices are at (±3,0)

4. Calculate the foci using the relation c^2=b^2−a^2

c^2=16−9

c^2=7

c=√(,7)

The foci are located at (0,±√(,7))

5. Sketch the graph by plotting the center (0,0) the vertices (0,4) and (0,−4) and the co-vertices (3,0) and (−3,0) then drawing a smooth curve through these points.

Final Answer

(x^2)/9+(y^2)/16=1* is a vertical ellipse with center *(0,0), vertices *(0,±4), and co-vertices *(±3,0)

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