The Locus - Q :D (1 Viewer)

[Smile12345]

Hello All.

I'm just not getting this question... Could you please help me?

Q. Find the equation of the locus of a point that moves so that it is equidistant from the point (1,-1) and the line y=3.

Thanks heaps in advance.

 

[asianese]

Let P be such a point with coordinates x and y.

If it's distance from (1,-1) equals the distance from y=3...then...

_________ = __________

(What's the distance between two general points?)

(What's the distance between a point and y=3?)

 

[braintic]

Since the equation says 'find' and not 'derive', just use one of the standard forms for a parabola (you should know that this locus satisfies the definition of a parabola).
The given point is the focus, the line is the directrix.
The vertex is halfway between the focus and directrix, ie. (1,1). So the focal length is 2. A diagram reveals that the parabola must be concave down.
So the equation is (x-1)^2 = 8(y-1).

 

[Smile12345]

Thanks Asianese... I managed to work I out!

I had one point as (1,-3) instead of (x,-3)!!

 

[Smile12345]

Since the equation says 'find' and not 'derive', just use one of the standard forms for a parabola (you should know that this locus satisfies the definition of a parabola).
The given point is the focus, the line is the directrix.
The vertex is halfway between the focus and directrix, ie. (1,1). So the focal length is 2. A diagram reveals that the parabola must be concave down.
So the equation is (x-1)^2 = 8(y-1).

Thanks heaps... Yeah, that's the way to do it... Was in an Exercise before I was meant to know all of the above!

 

[Smile12345]

I've got a couple of other questions...

Q. Given two points A(3,-2) and B(-1,7), find the equation of the locus of P(x,y) if the gradient of PA is twice the gradient of PB.

Q. Find the equation of the locus of a point that moves so that it is equidistant from the line 4x - 3y + 2 = 0 and the line 3x - 4y -7 = 0...

Thanks in advance for your help/hints...

 

[braintic]

I've got a couple of other questions...

Q. Given two points A(3,-2) and B(-1,7), find the equation of the locus of P(x,y) if the gradient of PA is twice the gradient of PB.

Q. Find the equation of the locus of a point that moves so that it is equidistant from the line 4x - 3y + 2 = 0 and the line 3x - 4y -7 = 0...

Thanks in advance for your help/hints...

(1)
m_PA = (y+2)/(x-3)
m_PB = (y-7)/(x+1)

So (y+2)/(x-3) = 2(y-7)/(x+1)

Then cross-multiply and simplify.

(2)
Let the point be P(x,y). Then use perpendicular distance formula twice, once for each line, setting the results equal to each other.
The only problem should be in dealing with the absolute values.
But if you know how to solve equations like |3x-1| = |x+2|, then it shouldn't be a problem.

 

[Smile12345]

For Q1.... I'm not getting it.... I've tried to cross multiply etc... I can't get the right answer.

 

[Smile12345]

Hmmmm.... Sorry I need Q2 - explained more too... Yes, I do know how to solve equations like that...

Thanks

 

[braintic]

For Q1.... I'm not getting it.... I've tried to cross multiply etc... I can't get the right answer.

Are you sure that the question mentioned gradient? I have never seen this question before, and it is actually a pretty pointless question.

Are you sure it wasn't a distance relation? ie. PA=2PB

 

[Smile12345]

Yes, the question did mention Gradient. I've just double checked it... Yes, I agree it is...

That's the way the question was worded...

 

[Smile12345]

Can someone please help??