Find the Properties (x^2)/16+(y^2)/12=1

Problem

(x^2)/16+(y^2)/12=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 a plus sign and a^2≠b^2 it is an ellipse centered at (0,0)

2. Determine the semi-major and semi-minor axes. Comparing the equation to the standard form, a^2=16 and b^2=12 Taking the square roots gives a=4 and b=√(,12)=2√(,3) Since a>b the major axis is horizontal along the x-axis.

3. Calculate the focal distance c using the relationship c^2=a^2−b^2

c^2=16−12

c^2=4

c=2

4. Find the vertices and co-vertices. The vertices are at (±a,0) which are (±4,0) The co-vertices are at (0,±b) which are (0,±2√(,3))

5. Locate the foci. The foci are at (±c,0) which are (±2,0)

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

e=2/4

e=0.5

Final Answer

Center: *(0,0), Vertices: *(±4,0), Foci: *(±2,0), Eccentricity: *0.5

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