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

Problem

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

Solution

1. Identify the type of conic section. Since the equation is in the form (x2)/(a2)+(y2)/(b2)=1 with positive coefficients, it is an ellipse centered at the origin (0,0)

2. Determine the lengths of the semi-axes. Here a2=16 and b2=9 which means a=4 and b=3

3. Locate the vertices on the x-axis. Since a=4 is under the x2 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 y2 term, the vertical co-vertices are at (0,3) and (0,−3)

5. Calculate the foci using the formula c2=a2−b2

c2=16−9

c2=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

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

Want more problems? Check here!