SOLUTION: At this point, you are now ready to take the summative assessment for learning plan 3. Place your answers on a whole sheet of paper. Show your complete solution. (20 points) 1.

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: At this point, you are now ready to take the summative assessment for learning plan 3. Place your answers on a whole sheet of paper. Show your complete solution. (20 points) 1.       Log On


   

Question 1169047: At this point, you are now ready to take the summative assessment for learning plan 3. Place your answers on a whole sheet of paper. Show your complete solution. (20 points)
1. Write the standard form of the equation of the hyperbola with the following characteristics:
Center at (2,3)
Vertices at (-1,3) and (5,3)
Covertices at (2,-2) and (2,8)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
recal:
Coordinates of the center: (h, k).
Coordinates of vertices: (h%2Ba, k) and (h+-+a,k)
Co-vertices correspond to b, the ” minor semi-axis length”, and coordinates of co-vertices:
(h,k%2Bb) and (h,k-b).
Foci have coordinates (h%2Bc,k) and (h-c,k). The value of c is given as, c%5E2+=+a%5E2+%2B+b%5E2.
equation:
%28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1

given:
Center at (2,3)=(h,k)
Thus, h=2, k=3.
Vertices at (-1,3) and (5,3)
The distance between the vertices is 2a. We can find this distance by subtracting the x-coordinates of the vertices:
2a=5-%28-1%29=6
a=3

so far your equation is:
%28x-2%29%5E2%2F3%5E2-%28y-3%29%5E2%2Fb%5E2=1

Covertices at (2,-2)and (2,8)
since covertices at (h,k%2Bb) and (h,k-b), we have
(2,3%2Bb) =(2,8) => 3%2Bb=8=> b=8-3 =>b=5

so, your equation is:
%28x-2%29%5E2%2F3%5E2-%28y-3%29%5E2%2F5%5E2=1
%28x-2%29%5E2%2F9-%28y-3%29%5E2%2F25=1