What curve does the equation #(x-3)^2/4+(y-4)^2/9=1# represent and what are its points of intersection with the axes ?

1 Answer
Sep 7, 2017

Answer:

This is an ellipse that does not intersect the axes...

Explanation:

Given:

#(x-3)^2/4+(y-4)^2/9=1#

Let's reduce the number of fractions we need to work with by multiplying both sides by #36# first to get:

#9(x-3)^2+4(y-4)^2=36#

Subtracting #36# from both sides and transposing, we get:

#0 = 9(x-3)^2+4(y-4)^2-36#

#color(white)(0) = 9(x^2-6x+9)+4(y^2-8y+16)-36#

#color(white)(0) = 9x^2-54x+81+4y^2-32y+64-36#

#color(white)(0) = 9x^2+4y^2-54x-32y+109#

We can find the intercepts with the #x# axis by substituting #y=0#, or equivalently covering up the terms involving #y# to find:

#0 = 9x^2-54x+109#

#color(white)(0) = (3x)^2-2(3x)(9)+81+28#

#color(white)(0) = (3x-9)^2+28#

This has no real solutions, so there are no intercepts with the #x# axis#.

We can find the intercepts with the #y# axis by substituting #x=0#, or equaivalently covering up the terms involving #x# to find:

#0 = 4y^2-32y+109#

#color(white)(0) = (2y)^2-2(2y)(8)+64+45#

#color(white)(0) = (2y-8)^2+45#

This has no real solutions, so there are no intercepts with the #y# axis.

Alternatively, we could have saved ourselves much of this algebra by noting that the equation:

#(x-3)^2/4+(y-4)^2/9=1#

is the standard form of the equation of an ellipse:

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

with centre #(h, k) = (3, 4)#, semi minor axis of length #a=2# (in the #x# direction) and semi major axis of length #b=3# (in the #y# direction).

So the ellipse is #1# unit from both axes...
graph{(x-3)^2/4+(y-4)^2/9=1 [-9, 11, -2.24, 7.76]}