Help with locus qn! (1 Viewer)

[Getteral09]

hey can anyone work out this maths question?Find the equation of the focus of a point p(x, y) that is subject to the following conditions: The distance of a point from the line x= -3 is two-fifths of its distance from the line y=-1." So how am i meant to find the equation?
Thanks,Getteral

 

[SoulSearcher]

What are the answers as given in the book?

 

[Getteral09]

The answers at the back of Fitzpatrick have two answers. They are 5x-2y+13=0 which OneByTwo calculated right while the other anwser is 5x+2y+17=0.



So how do you get that equation: 5x+2y+17=0?

thanks so much,

Getteral

 

[SoulSearcher]

Well there are actually two solutions to the question, this can be seen by drawing a graph.
You notice that there are two lines which fulfil that equation, one has gradient 5/2 which gives the first answer by using point-gradient formula with the point (-3,-1).
The second line has gradient -5/2, which, when used with the point (-3,-1), gives the second equation of the line.

 

[zeek]

You can also use the distance formula where Px=(2/5)Py in this case

 

[Getteral09]

hey how did you get the gradient to be 5/2.

 

[Riviet]

By a definition of absolute value, sqrt(x²)=|x|

Using the distance formula,

sqrt{(x-(-3))²}=2/5.sqrt{(y-(-1))²}

5|x+3|=2|y+1|

We consider the two cases when both sides are positive and the two sides are of opposite sign:

1) 5(x+3)=2(y+1)
5x+15-2y-2=0
5x-2y+13=0

2) 5(x+3)=-2(y+1)
5x+15+2y+2=0
5x+2y+17=0