# Find the center and foci of the ellipse. (x+1)^2/28 + (y+2)^2/64 = 1 Find the standard form of the equation of the hyperbola with the given characteristics. vertices (2,1)(6,1) asymptote y=5/2x-9...

Find the center and foci of the ellipse. (x+1)^2/28 + (y+2)^2/64 = 1

Find the standard form of the equation of the hyperbola with the given characteristics. vertices (2,1)(6,1) asymptote y=5/2x-9 and y=-5/2x+11

justaguide | Certified Educator

The standard equation of an ellipse with center (h,k) is `(x - h)^2/a^2 + (y - k)^2/b^2 = 1` with foci c where `c^2 = a^2 + b^2`

For the ellipse `(x+1)^2/28 +(y+2)^2/64 = 1` , the center is (-1, -2) and the foci is `sqrt(28+64) = sqrt 92`

The vertices of the hyperbola are (2, 1) and (6, 1). The equation of a hyperbola opening towards the sides is `(x - h)^2/a^2 - (y - k)^2/b^2 = 1` where the asymptotes are `+-(b/a)(x - h) + k`

Here, the asymptotes are y = 2.5x - 9 and y = -2.5x + 11

This can be written as: y = 2.5*(x - 4) + 1 and y = -2.5*(x - 4) + 1

h = 4, k = 1, (b/a) = 2.5, b + a = 4

b = 2.5a , 2.5a + a = 4

3.5a = 4

a = 8/7, b = 20/7

The equation of the hyperbola is `(49*(x - 4)^2)/64 - (49*(y - 1)^2)/(400) = 1`