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 ?

Question

Avery Alexander · Accepted Answer

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

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]}

Abigail Allen · Answer

undefined The equation represents an ellipse centered at (3, 4) with a horizontal axis length of 4 units and a vertical axis length of 6 units.

To find the points of intersection with the axes:
1. For the x-axis: Set $y = 0$ and solve for $x$.
2. For the y-axis: Set $x = 0$ and solve for $y$.

For the x-axis:
$$
\frac{{(x - 3)^2}}{4} + \frac{{(0 - 4)^2}}{9} = 1
$$
$$
\frac{{(x - 3)^2}}{4} + \frac{16}{9} = 1
$$
$$
\frac{{(x - 3)^2}}{4} = 1 - \frac{16}{9}
$$
$$
\frac{{(x - 3)^2}}{4} = \frac{9}{9} - \frac{16}{9}
$$
$$
\frac{{(x - 3)^2}}{4} = \frac{{-7}}{9}
$$
There are no real solutions since the right-hand side is negative.

For the y-axis:
$$
\frac{{(0 - 3)^2}}{4} + \frac{{(y - 4)^2}}{9} = 1
$$
$$
\frac{9}{4} + \frac{{(y - 4)^2}}{9} = 1
$$
$$
\frac{{(y - 4)^2}}{9} = 1 - \frac{9}{4}
$$
$$
\frac{{(y - 4)^2}}{9} = \frac{4}{4} - \frac{9}{4}
$$
$$
\frac{{(y - 4)^2}}{9} = \frac{{-5}}{4}
$$
There are no real solutions since the right-hand side is negative.

So, the ellipse does not intersect the x-axis or the y-axis in real numbers.