Question

Vertices (0,8) and (0,-8) asymptotes y= ±2x find the equation

Answer 1

The equation of a hyperbola with vertices at (0,8) and (0,-8) and asymptotes y=± 2x is (y²/64) - (x²/256) = 1.

Step-by-step explanation:

The student is asking for help to find the equation of a hyperbola given its vertices and asymptotes. Since the vertices are (0,8) and (0,-8), we know that the center of the hyperbola is at the origin (0,0) and the transverse axis is along the y-axis.

The distance between the vertices and the center is 8 units, so the length of the transverse axis (2a) is 16, giving us a value of a=8. The asymptotes of a hyperbola are given by y=±(b/a)x for a hyperbola that opens up and down. Given the equation of the asymptotes y=± 2x, we find that b/a = 2.

With a known, we can solve this to find b=16. The standard form of the equation for a vertical hyperbola centered at the origin is (y²/a²) - (x²/b²) = 1. Plugging in the values for a and b, we get (y²/64) - (x²/256) = 1, which is the desired equation.

Answer 2

The equation of the parabola is y = -2x^2 + 8.

Step-by-step explanation:

To find the equation of the parabola with vertices (0,8) and (0,-8) and asymptotes y= ±2x, we can start by considering the general equation of a parabola: y = ax^2 + bx + c. Since the parabola has symmetry along the y-axis, the equation can be simplified to y = ax^2 + c.

Using the given information, we know that the parabola passes through the points (0,8) and (0,-8), so we can substitute these coordinates into the equation. This gives us the system of equations 8 = c and -8 = c. Solving this system, we find that c = 8 and a = -2. Therefore, the equation of the parabola is y = -2x^2 + 8.