Find the Properties 16x^2+9y^2=1

Problem

16*x2+9*y2=1

Solution

1. Identify the type of conic section by observing that both x2 and y2 terms are positive and have different coefficients, which indicates an ellipse.

2. Rewrite the equation in standard form (x2)/(a2)+(y2)/(b2)=1 by expressing the coefficients as denominators.

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

3. Determine the center (h,k) by noting there are no horizontal or vertical shifts.

(h,k)=(0,0)

4. Identify the semi-axes a and b Since 1/9>1/16 the ellipse is vertical with b2=1/9 and a2=1/16

b=√(,1/9)=1/3

a=√(,1/16)=1/4

5. Calculate the distance to the foci c using the relation c2=b2−a2 for a vertical ellipse.

c2=1/9−1/16

c2=(16−9)/144=7/144

c=√(,7)/12

6. Find the vertices and co-vertices. The vertices are at (h,k±b) and the co-vertices are at (h±a,k)

Vertices=(0,±1/3)

Co-vertices=(±1/4,0)

7. Find the foci using the coordinates (h,k±c)

Foci=(0,±√(,7)/12)

8. Determine the eccentricity e using the formula e=c/b

e=√(,7)/12/1/3=√(,7)/4

Final Answer

Center: *(0,0), Vertices: *(0,±1/3), Foci: *(0,±√(,7)/12), Eccentricity: √(,7)/4

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