Graph (x^2)/16+(y^2)/6=1

Problem

(x^2)/16+(y^2)/6=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. Here, a^2=16 and b^2=6 Taking the square roots gives a=4 and b=√(,6)≈2.45

3. Locate the vertices on the major axis. Since a^2>b^2 the major axis is horizontal. The vertices are located at (±a,0) which are (4,0) and (−4,0)

4. Locate the co-vertices on the minor axis. The co-vertices are located at (0,±b) which are (0,√(,6)) and (0,−√(,6))

5. Calculate the foci using the relation c^2=a^2−b^2 Substituting the values gives c^2=16−6=10 so c=√(,10)≈3.16 The foci are at (±√(,10),0)

6. Sketch the graph by plotting the center, vertices, and co-vertices, then drawing a smooth curve to connect them into an oval shape.

Final Answer

(x^2)/16+(y^2)/6=1* is an ellipse with vertices *(±4,0)* and co-vertices *(0,±√(,6))

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