Question

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 ?

Answer 1

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]}