Question

How do you do this: Find the area enclosed by the locus of all points that are exactly 1/3 as far away from the point (1,2) as they are from the line y=-2x+24?

I know this is talking about an ellipse using the definition of foci and directrix but I don't know where to start since this is not in standard position.

Answer 1

`(x-1)^2+(y-2)^2=80/9`

Explanation:

Much of the key to solving this problem is in drawing what you know. For example, you are given the line `y=-2x+24` and the point `(1,2)`. Drawing those on the same graph gives.

We are told that something ("the locus of all points") is one third as far the distance between the point `(1,2)` and the line `y=-2x+24`. So let's find the distance between the point `(1,2)` and the line `y=-2x+24`.

It would be perpendicular and, thus, have a slope that is perpendicular to the line, as well as pass through `(1,2)`.

`m_1=1/2` (opposite sign and inverse fraction of `m=-2`)

`y-y_1=m_1(x-x_1)`

`y-2=1/2(x-1)`

`y=1/2x+3/2`

The distance formula between `(1,2)` and the point were the two lines intersect at `(9,6)` is

`d=sqrt((9-1)^2+(6-2)^2)`

`d=sqrt(8^2+4^2)`

`d=sqrt(80)`

One third that distance is simply

`d/3=(sqrt(80))/3`

The way I'm reading the question, the "area enclosed by all points" that have exactly the same distance from the center (i.e., one third the distance from center to line), is a circle.

`(x-1)^2+(y-2)^2=(sqrt(80)/(3))^2`