How do you write an equation of a ellipse with vertices (0,2), (4,2), and endpoints of the minor axis (2,3), (2,1)?

1 Answer
Nov 29, 2017

Answer:

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

Explanation:

Here we have an ellipse with vertices at #(0,2) and (4,2)# and the minor axis between #(2,3) and (2,1)#

The equation of an ellipse centred at the origin with major axis length #(2a)# and minor axis length #(2b)# is:

#x^2/a^2 + y^2/b^2 =1#

Here we have an ellipse with major axis on #y=2#
(Since the vertices are both on #y=2#)

And minor axis on #x=2#
(Since the endpoints of the minor axis are both on #x=2#)

Hence, the centre of the ellipse is at #(2,2)#

From the above we can calculate the length of the major axis to be:
#4-0 = 4 #

#-> a=4/2 =2#

and the length of the minor axis to be:
#3-1 =2#

#-> b= 2/2 =1#

Applying the shifts of the centre from #(0,0)# to #(2,2)# our standard equation becomes:

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

Substituting for #a and b# above

#-> (x-2)^2/2^2 + (y-2)^2/1^2 =1#

Hence, the equation of our ellipse is: #(x-2)^2/4 + (y-2)^2/1 = 1#

We can see this ellipse in the graphic below that satisfies the given conditions.

graph{(x-2)^2/4+(y-2)^2/1=1 [-3.26, 7.84, -0.51, 5.04]}