Locus and parabola help (1 Viewer)

[poopoohead]

Can someone help me with this question please:

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

what is the best way to this? im just a bit confused

The answer is x-7y+9=0, and 7x + y - 5= 0. Also, Why are there two answers?

I have 2 more quick questions- these ones are probably really easy ones-

1. Find the equation of the parabola with the coordinates of the vertex (0,0), the equation of the axis x=0 and focal length 8. The answer is x²= 32y and x²= -32y. Why is there is two answers????????????

2. Find the equation of the parabola with coordinates of the vertex (0,0) and the equation of the axis x=0, and passing through the point (-8, 2). The answer is x² = 32y

thank you for your help!

 

[pLuvia]

1)
(x-h)²=4|a|(y-k)
Vertex at (0,0)=(h,k)
x²=4|a|y
Focal length=32=4|a|
|a|=8
a=8 or -8
x²=32y
or
x²=-32y

Since "a" is a distance it must be positive so it has the absolute value signs on it

2)
(x-h)²=4a(y-k)
Vertex at (0,0)=(h,k)
x²=4ay
Passes through (-8,2), sub these into the eqn
(64)=8a
a=8
x²=32y

 

[Riviet]

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

what is the best way to this? im just a bit confused

The answer is x-7y+9=0, and 7x + y - 5= 0. Also, Why are there two answers?

There are two answers because there are two possible lines that can be equidistant from the two given lines in the question.

Try this:
Draw a number plane about 1/3 the size of an A4 piece of paper. Draw two random lines that are not parallel to each other and form a random X. Get two pens and put one over the other so they form a + shape. Get this + shape and put it on the X so that each pen is equidistant from each diagonal of the X. The two pens represent the two possible lines that are equidistant from the two original lines.

Let a random point on the locus be P(x,y).
For 4x-3y+2=0
distance = |ax+by+c|/rt(a²+b²)
=|4x-3y+2|/rt(4²+3^2[)
=|4x-3y+2|/5

For 3x+4y-7=0,
distance = |ax+by+c|/rt(a²+b²)
=|3x+4y-7|/rt(3²+4²)
=|3x+4y-7|/5

Since the distances to both lines are equal, we equate the distances:
|4x-3y+2|/5 = |3x+4y-7|/5
There are two cases:
1) Both sides have the same sign (ie both sides are positive or both sides are negative)
2) One side is positive and the other negative

Case 1:
(4x-3y+2)/5 = (3x+4y-7)/5
Bring everything to the left side,
x-7y+9=0

Case 2:
-(4x-3y+2)/5 = (3x+4y-7)/5
-4x+3y-2 = 3x+4y-7
Bring everything to the right side,
7x+y-5=0

 

[Shady01]

ahh locus feelin tha pain feelin tha pain

 

[poopoohead]

thankyou . yeah whole hsc giving me pain in the brain