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

Problem

(x^2)/16+(y^2)/9=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 coefficients, 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=9 which means a=4 and b=3

3. Locate the vertices on the x-axis. Since a=4 is under the x^2 term, the horizontal vertices are at (4,0) and (−4,0)

4. Locate the co-vertices on the y-axis. Since b=3 is under the y^2 term, the vertical co-vertices are at (0,3) and (0,−3)

5. Calculate the foci using the formula c^2=a^2−b^2

c^2=16−9

c^2=7

c=√(,7)

6. Sketch the graph by drawing a smooth curve through the four points (4,0) (−4,0) (0,3) and (0,−3)

Final Answer

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

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