Find the Properties ((y-2)^2)/16-((x+1)^2)/144=1

Problem

((y−2)2)/16−((x+1)2)/144=1

Solution

1. Identify the type of conic section. Since the equation is in the form ((y−k)2)/(a2)−((x−h)2)/(b2)=1 it is a vertical hyperbola.

2. Determine the center (h,k) by looking at the terms (x+1) and (y−2) The center is (−1,2)

3. Calculate the semi-transverse axis a and semi-conjugate axis b Here, a2=16 so a=4 and b2=144 so b=12

4. Find the distance from the center to the foci c using the relation c2=a2+b2

c2=16+144=160

c=√(,160)=4√(,10)

5. Locate the vertices by moving a units vertically from the center.

Vertices: *(−1,2±4)

Vertices: *(−1,6),(−1,−2)

6. Locate the foci by moving c units vertically from the center.

Foci: *(−1,2±4√(,10))

7. Determine the equations of the asymptotes using the formula y−k=±a/b*(x−h)

y−2=±4/12*(x+1)

y−2=±1/3*(x+1)

8. Calculate the eccentricity e using e=c/a

e=(4√(,10))/4=√(,10)

Final Answer

Center: *(−1,2), Vertices: *(−1,6),(−1,−2), Foci: *(−1,2±4√(,10)), Asymptotes: *y−2=±1/3*(x+1)

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