Find the Properties ((x+3)^2)/144-((y-2)^2)/25=1

Problem

((x+3)^2)/144−((y−2)^2)/25=1

Solution

1. Identify the conic section by observing the equation is in the form ((x−h)^2)/(a^2)−((y−k)^2)/(b^2)=1 which represents a horizontal hyperbola.

2. Determine the center (h,k) by extracting the values from the terms (x+3) and (y−2)

h=−3

k=2

Center=(−3,2)

3. Calculate the semi-axes a and b by taking the square roots of the denominators.

a^2=144⇒a=12

b^2=25⇒b=5

4. Find the focal distance c using the relationship c^2=a^2+b^2

c^2=144+25=169

c=√(,169)=13

5. Locate the vertices by moving a units to the left and right of the center.

Vertices=(h±a,k)

Vertex 1=(−3+12,2)=(9,2)

Vertex 2=(−3−12,2)=(−15,2)

6. Locate the foci by moving c units to the left and right of the center.

Foci=(h±c,k)

Focus 1=(−3+13,2)=(10,2)

Focus 2=(−3−13,2)=(−16,2)

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

y−2=±5/12*(x+3)

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

e=13/12

Final Answer

Center: *(−3,2), Vertices: *(9,2),(−15,2), Foci: *(10,2),(−16,2), Asymptotes: *y−2=±5/12*(x+3)

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