Find the Properties 3x^2+4y^2=12

Problem

3*x^2+4*y^2=12

Solution

1. Convert to standard form by dividing both sides of the equation by 12 to set the right side equal to 1

(3*x^2)/12+(4*y^2)/12=12/12

2. Simplify the fractions to identify the denominators a^2 and b^2

(x^2)/4+(y^2)/3=1

3. Identify the conic section as an ellipse centered at the origin (0,0) because the equation is in the form (x^2)/(a^2)+(y^2)/(b^2)=1 with a^2>b^2

Type: Horizontal Ellipse

4. Determine the semi-axes by taking the square roots of the denominators.

a=√(,4)=2

b=√(,3)

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

c^2=4−3=1

c=1

6. Identify the vertices along the major axis (x-axis) using (±a,0)

Vertices: *(2,0),(−2,0)

7. Identify the co-vertices along the minor axis (y-axis) using (0,±b)

Co-vertices: *(0,√(,3)),(0,−√(,3))

8. Identify the foci along the major axis using (±c,0)

Foci: *(1,0),(−1,0)

9. Calculate the eccentricity using the formula e=c/a

e=1/2

Final Answer

Center: *(0,0), Vertices: *(±2,0), Foci: *(±1,0), Eccentricity: 1/2

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