Find the equation of the ellipse with the following conditions below. Explain completely the process of arriving at your answer.

Question

a. center: (1, 4); a = 2c; vertices: (1, 0), (1, 8)

d. vertices: (5, 0), (5, 12); endpoints of the minor axis: (1, 6), (9, 6) 

e. foci at (0 , ±3) ; x-intercepts are ±2

Answer 1

it helps to draw a sketch of the 3 ellipses, their points and focui

a.)

(x-h)/b^2 + (y-k)^2/a^2 = 1 is the general formula with (h,k)=center

2a = major axis, 2b=minor axis

(x-1)^2/20 +(y-4)^2/(16) =1

4(x-1)^2 +5(y-4)^2 = 80

endpoints or vertices: (1,0),(1,8) on the major axis

a=4=2c, c=2

b^2= c^2 +a^2 = 4+16=20

b=sqr20 =2sqr5

major axis = 2a= 8

with minor axis=2b=4sqr5=2sqr20

with center (1,4)

d)

center (5,6)

vertices (5,0),(5,12), (1,6),(9,6)

major axis =2a =12, a=6

minor axis =2b=9-1=8, b=4

(x-h)^2/b^2 + (y-k)^2/a^2 = 1

(x-5)^2/16 +(x-6)^2/36 = 1

9(x-5)^2 + 4(x-16)^2 = 144

e.)

foci (0,+/-3) and x intercepts (+/-2,0)= vertices (-2,0),(2,0)

center is the origin

minor axis =2b=4, b=2

major axis =2a= 2sqr13, a=sqr13, vertices (-sqr13,0),(sqr13,0)

c^2=a^2 -b^2

a^2 =9+4

a =sqr13

(x-h)^2/b^2 + (y-k)^2/a^2 = 1, (h,k)=(0,0), b=2, a=sqr13

x^2/4 +y^2/13 = 1

13x^2 +4y^2 = 52