Question

1.     The general equation of a conic section is given by -4x^2 + 25y^2 -50y +125 =0.

(a)   Identify what conic section is represented by the equation above. Show all your steps in handwritten work and insert an image of the work below.

(b)   Write the equation of this conic section in graphing form. Show all your steps in handwritten work and insert an image of the work below.

(c)   Find the coordinates of the center, the foci, and the vertices. Show all your steps in handwritten work and insert an image of the work below.

(d)   Write the equations of the lines of the asymptotes. Show all your steps in handwritten work and insert an image of the work below.

(e)   Graph the conic section. Label the center, the foci, the vertices and the asymptotes. Show all your steps in handwritten work and insert an image of the work below. There is a graph on the next page you may use.

Answer 1

-4x^2 + 25y^2 - 50y + 125 =0

a) as the x^2 and y^2 coeffinents are different and they have opposite signs

the conic section is hyperbola

b) equation of conic section in graphing form

-4x^2 + 25 y^2 - 50y + 125 = 0

x^2/ 25 - (y-1)^2/4 = 1

c) center = (0,1)

c^2 = 5^2 + 2^2

c = sqrt 29

foci = ( sqrt 29 , 1 ) , ( - sqrt 29, 1 )

vertices = (5,1) , (-5,1 )

d) equation of asymptotes are y = +- b/a ( x- h ) + k

y = 2x/5 + 1 , y = -2x/5 + 1

e) graph shoen below