Question

Write the equation of a hyperbola with vertices (3, -1) and (3, -9) and co-vertices (-6. -5) and (12, -5).

Answer 1

@ivanmlerner

Answer 2

Try using the same method I used before, remember what a and b are (drawing), but since this time the hyperbola is not centered in the origin, what would you need to do to the formula to change the center? Try answering that, because it gets easier to understand.

Answer 3

basically trying to bring the letters together to get to the center?

Answer 4

I'm not sure I understood what you said, but a and b are independent of the center, the change is in another part of the equation, think of x and y as distances between the point in the hyperbola and the center, if you have a hyperbola that is not centered in the origin, you'd have to add or subtract something from x and y to keep the distance the same even thought you are moving the hyperbola.

Answer 5

Draw the points also, because it gets really easy to see the center. Or look at the coordinates that remain constant in a line in the points given (3, -1) (3, -9) all have x=3 and (-6, -5) (12, -5) have y=-5

Answer 6

so how would i for an equation from this?

Answer 7

Ok, the center is (3, -5), a is (12-(-6))/2=9, that is also easy to see if you draw the points, b is (-1-(-9))/2=4.
The equation for the hyperbola centered on the origin would be:\[\left( \frac{ x }{ 9 } \right)^2+\left( \frac{ y }{ 4 } \right)^2=1\]But you want it to be centered at (3, -5), and that means that when y=0, x needs to be 3+-a, and when x=0, y needs to be -5+-b because we would be on the axis that cross the hyperbola. So soving the equation centered on the origin for each case we see that x needs to be replaced by x-3 and y by y+12.\[\left( \frac{ x-3 }{ 9 } \right)^2+\left( \frac{ y+12 }{ 4 } \right)^2=1\]

Answer 8

thankss!