Question

What are the foci of the hyperbola represented by the equation (x^2)/64 - (y^2)/32 =1

Answer 1

x = √2y^2 + 64

Explanation:

= 1/64 x^2 + -1/32 y^2 + 1/32 y^2 = 1 + 1/32 y^2

= 1/64 x^2/ 1/64 = 1/32 y^2 + 1/ 1/64

= x^2 = 2y^2 + 64

= x = √2y^2 + 64

Hope this helps!

Answer 2

x = √2y^2 + 64 or x = √2y^2 + 64

Explanation:

Solve for x.

x^2/64 - y^2/32 = 1

Step 1: Add 1/32 y^2 to both sides.

1/64 x^2 + -1/32 y^2 + 1/32 y^2 = 1 + 1/32 y^2

Step 2: Divide both sides by 1/64.

1/64 x^2/ 1/64 = 1/32 y^2 + 1/ 1/64

x^2 = 2y^2 + 64

Step 3:

x = √2y^2 + 64 or x = √2y^2 + 64