What are the vertices of the hyperbola given by the equation (y^2/81)-(x^2/49) = 1?

(±7, 0)

(±9, 0)

(0, ±7)

(0, ±9)

Question

What are the vertices of the hyperbola given by the equation (y^2/81)-(x^2/49)  = 1?

 (±7, 0)

 (±9, 0)

 (0, ±7)

 (0, ±9)

valpey · Answer

When y is zero there are no solutions to the equation. When x is zero there are two solutions to the equation. The vertices of a Hyperbola are particular solutions to the equation (the are on the graph of hyperbola).

anonymous · Answer

so what is the answer?

anonymous · Answer

The point is that
When y is zero there are no solutions to the equation.
Since this hyperbola doesn't have xy terms it will open "east-west"

anonymous · Answer

sorry north-south

anonymous · Answer

attachment

anonymous · Answer

The actual values for the vertices are solutions when x = 0

anonymous · Answer

I still dont know the answer..

anonymous · Answer

Do you see the north-south part?

anonymous · Answer

y cannot be equal to 0, b/c then we have
-(x^2/49)  = 1

anonymous · Answer

and since x^2 > 0 always, this has no solutions.

anonymous · Answer

Think about this
The actual values for the vertices are solutions when x = 0