Graph ((x-0)^2)/(5^2)+((y-0)^2)/(3^2)=1

Problem

((x−0)^2)/5^2+((y−0)^2)/3^2=1

Solution

1. Identify the type of conic section. The equation is in the standard form of an ellipse centered at (h,k) which is ((x−h)^2)/(a^2)+((y−k)^2)/(b^2)=1

2. Determine the center (h,k) By comparing the given equation to the standard form, the center is (0,0)

3. Find the lengths of the semi-axes. The value a^2=5^2 gives a=5 (horizontal semi-major axis), and b^2=3^2 gives b=3 (vertical semi-minor axis).

4. Locate the vertices and co-vertices. The vertices are at (h±a,k) which are (5,0) and (−5,0) The co-vertices are at (h,k±b) which are (0,3) and (0,−3)

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

c^2=25−9

c^2=16

c=4

The foci are located at (±4,0)

6. Sketch the graph by plotting the center, vertices, and co-vertices, then drawing a smooth curve through the points to form the ellipse.

Final Answer

((x−0)^2)/5^2+((y−0)^2)/3^2=1* is an ellipse centered at *(0,0)* with vertices *(±5,0)* and co-vertices *(0,±3)

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