write an equation of the conic section:

a hyperbola with vertices (0,-6) and (0,6) and asymptotes y=(3/4)x and y=(-3/4)x

nvm, i figured this one out

To find the equation of a hyperbola with the given vertices and asymptotes, we can start by considering the general form of a hyperbola centered at the origin:

x²/a² - y²/b² = 1

First, let's find the value of 'a' which represents the distance from the center to the vertices. In this case, the vertices are at (0, -6) and (0, 6), so the value of 'a' is the distance between these points, which is 6 units. Therefore, a = 6.

Next, let's find the value of 'b', which represents the distance from the center to the asymptotes. The slope of the asymptotes is given as 3/4 and -3/4, which means the slopes of the asymptotes are equal to b/a and -b/a, respectively. Thus, b/a = 3/4.

To find 'b', we can substitute a = 6 into the equation b/a = 3/4 and solve for b:

b/6 = 3/4

Cross-multiplying:

4b = 18

Dividing by 4:

b = 18/4 or b = 9/2

Now, we have the values of 'a' and 'b', so we can write the equation of the hyperbola:

x²/6² - y²/(9/2)² = 1

Simplifying:

x²/36 - y²/(81/4) = 1

To simplify further, we can multiply both sides of the equation by 36 to remove the denominators:

36(x²/36) - 36(y²/(81/4)) = 36

x² - 4y²/81 = 36

Finally, we can multiply both sides by 81 to eliminate the fractional coefficient of y²:

81x² - 4y² = 2916

So, the equation of the hyperbola is 81x² - 4y² = 2916.