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

Problem

16*x^2+9*y^2=1

Solution

1. Identify the type of conic section by observing that both x^2 and y^2 terms are positive and have different coefficients, which indicates an ellipse.

2. Rewrite the equation in standard form (x^2)/(a^2)+(y^2)/(b^2)=1 by expressing the coefficients as denominators.

(x^2)/1/16+(y^2)/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 b^2=1/9 and a^2=1/16

b=√(,1/9)=1/3

a=√(,1/16)=1/4

5. Calculate the distance to the foci c using the relation c^2=b^2−a^2 for a vertical ellipse.

c^2=1/9−1/16

c^2=(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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