Find the Properties (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)/(b2)+(y2)/(a2)=1 with a2>b2 it is a vertical ellipse centered at the origin (0,0)

2. Determine the values of a and b We have a2=16 so a=4 We have b2=9 so b=3

3. Calculate the distance to the foci c using the relationship c2=a2−b2

c2=16−9

c2=7

c=√(,7)

4. Find the vertices and co-vertices. The vertices are located at (0,±a) which are (0,4) and (0,−4) The co-vertices are located at (±b,0) which are (3,0) and (−3,0)

5. Locate the foci. The foci are located on the major axis at (0,±c) which are (0,√(,7)) and (0,−√(,7))

6. Compute the eccentricity e using the formula e=c/a

e=√(,7)/4

Final Answer

Center: *(0,0), Vertices: *(0,±4), Co-vertices: *(±3,0), Foci: *(0,±√(,7)), Eccentricity: √(,7)/4

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