233,778 views
2 votes
2 votes
I need help look at the image attached

I need help look at the image attached-example-1
User Uniqua
by
2.5k points

1 Answer

6 votes
6 votes

Answer:


((x-2)^2)/(9)-(y^2)/(4)=1

Explanation:

Given points:

  • (2±3√2, ±2)
  • (5, 0)
  • (-1, 0)

Standard equation of a horizontal hyperbola (opening left and right):


\boxed{((x-h)^2)/(a^2)-((y-k)^2)/(b^2)=1}

where:

  • center = (h, k)
  • vertices = (h±a, k)

Points (5, 0) and (-1, 0) are the vertices of the hyperbola.

The center (h, k) is midway between the vertices. Therefore:


\implies h=(5-1)/(2)=2


\implies k=0

Substitute the found values of h and k into the formula:


((x-2)^2)/(a^2)-(y^2)/(b^2)=1

Substitute point (5, 0) and solve for a²:


\implies ((5-2)^2)/(a^2)-(0^2)/(b^2)=1


\implies (3^2)/(a^2)=1


\implies (9)/(a^2)=1


\implies 9=a^2

Substitute one of the (2±3√2, ±2) points and the found value of a² and solve for b²:


\implies ((2+3√(2)-2)^2)/(9)-(2^2)/(b^2)=1


\implies ((3√(2))^2)/(9)-(4)/(b^2)=1


\implies (18)/(9)-(4)/(b^2)=1


\implies 2-(4)/(b^2)=1


\implies (4)/(b^2)=1


\implies 4=b^2

Substitute the found values of a² and b², together with the values of h and k, into the formula to create the equation of the hyperbola:


((x-2)^2)/(9)-(y^2)/(4)=1

I need help look at the image attached-example-1
User Esqarrouth
by
2.9k points