2 locus questions help (1 Viewer)

[bos1234]

Find the locus of a point P(x,y) which lies above the x-axis so that the sum of its distances to the origin and the x-axis is 2

If P(x,y) is any point on the line y = 4x+ 3, show that the midpt M of OP has coordinates (1/2x, 1/2(4x+3)). Hence find the locus of M

 

[vulgarfraction]

first part:

/sqrt{x^2+y^2} + |y| =2
/sqrt{x^2+y^2} =2 - |y|
x^2+y^2 =4 + y^2 - 2|y|
x^2 + 2 |y| - 4 = 0

but P is above x axis ==> |y| = y

so y = -(1/2)x^2 - 2

second part:

P(x,y) on y = 4x+3 is P(x, 4x+3) for given x.

Midpoint formula with O(0,0): M = (x/2, {4x+3}/2).

Locus of m: y = {4(2x)+3}/2 = 4x + 3/2.

Hope that helps. (I'm going to desperate measures to procrastinate here.)

 

[bos1234]

first part:

/sqrt{x^2+y^2} + |y| =2
/sqrt{x^2+y^2} =2 - |y|
x^2+y^2 =4 + y^2 - 2|y|
x^2 + 2 |y| - 4 = 0

but P is above x axis ==> |y| = y

so y = -(1/2)x^2 - 2

second part:

P(x,y) on y = 4x+3 is P(x, 4x+3) for given x.

Midpoint formula with O(0,0): M = (x/2, {4x+3}/2).

Locus of m: y = {4(2x)+3}/2 = 4x + 3/2.

Hope that helps. (I'm going to desperate measures to procrastinate here.)

Hello,

Thanks for fast reply

What was the condition for the locus to be this?

Locus of m: y = {4(2x)+3}/2 = 4x + 3/2.

 

[vulgarfraction]

Hello,

Thanks for fast reply

What was the condition for the locus to be this?

Locus of m: y = {4(2x)+3}/2 = 4x + 3/2.

Well, when x = x_0/2, y = {4x_0+3}/2 from previous step. Substitute x for x_0 in y, knowing that x_0 = 2x.