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

Problem

(x2)/9+(y2)/16=1

Solution

1. Identify the type of conic section. Since the equation is in the form (x2)/(a2)+(y2)/(b2)=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 x2 is a2=9 so a=3 The denominator under y2 is b2=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 c2=b2−a2

c2=16−9

c2=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

(x2)/9+(y2)/16=1* is a vertical ellipse with center *(0,0), vertices *(0,±4), and co-vertices *(±3,0)

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