locus/parabola questions (1 Viewer)

[Arithela]

1. S is the point (4,0) and d is the line x = 1. The point P(x,y) moves so that the distance of P to d is half the distane of P to S. Show the equation of the locus of P is 3x^2 - y^2 = 12. (This curve is a hyperbola).

2. A is a point where the circle with equation x^2 + y^2 = 16 cuts the X-axis. Find the locus of the midpoints of all chords of this circle that contain the point A.

3. Find the equation of the locus of the midpoints of all chords of length 4 units of the circle with equation x^2 + y^2 - 4x + 2y = 4.

4. A point P(x,y) moves so that its distance from (3,4) is proportional to its distance from (-1.2). Find the equation of the locus of P if the origin is a point on the locus.

5. What can be said about the centres of all circles that pass through the points (2,0) and (6,4)? What is the locus of the centres?

6. A ladder that is 6m long rests with one end on the horizontal ground, and the other end against a vertical wall. Considering the ground and the wall as the x and y axis respectively, find the locus of the midpoint of the ladder.

 

[108796]

q5: its a straight line perpendicular to the line between (2,0) and (6,4); and bisecting it (ie goes thru (4, 2).

ie equation of line is: y=-x+6

as a "proof", u that line above, then pick any point (x,y) on that line besides (4,2). draw the two lines from (x,y) to (2,0) and (x,y) to (6,4), and the line from (2,0) to (6,4). So u then get two triangles, which are congruent (right angle, one side common and two other sides are equal since the y=-x+6 bisects the other line).
therefore, the lines from (x,y) to (2,0) and (x,y) to (6,4) are equal
--> they are radii of the circle.

 

[108796]

q6

its the 1st quadrant of a circle centre (0,0) radius 3.

now, x^2+y^2=36 by phythagoras coz we got a ladder lying against a wall

then x^2/4+y^2/4 = 9
or (x/2)^2 +(y/2)^2=9
let x/2 = K, y/2 = L
and K^2 + L^2=9
now, the coordinates of the midpoint are (x/2,y/2) ie (K,L), so K^2+L^2=9 is the required locus.
this is indeed a circle centre (0,0) radius 3

this proof looks solid but im not too sure.

 

[108796]

4. A point P(x,y) moves so that its distance from (3,4) is proportional to its distance from (-1,2). Find the equation of the locus of P if the origin is a point on the locus.

distance formula (same as pythag):

sqrt[(x-3)^2+(y-4)^2]=k*sqrt[(x+1)^2+(y-2)^2] for some constant k, k is real

since it goes thru origin, P(0,0) satisfies this locus, so we can evaluate k:

sqrt(9+16)=ksqrt(1+4)
or k = 5/root 5= root 5

so:
sqrt[(x-3)^2+(y-4)^2]=(root 5)*sqrt[(x+1)^2+(y-2)^2]
squaring both sides and simplify,
0=4x^2+16x+4y^2-12y
ie. x^2+4x+y^2-3y=0
ie. x^2+4x+4 +y^2-3y+2.25=4+2.25 (completing the squares)
ie. (x+2)^2 + (y-1.5)^2 = 6.25

--> this locus is a circle centre (-2, 1.5), radius root 6.25 = 2.5

 

[108796]

3. Find the equation of the locus of the midpoints of all chords of length 4 units of the circle with equation x^2 + y^2 - 4x + 2y = 4.

the equation of the circle given: the squares can be completed to give:

(x-2)^2 + (y+1)^2 = 9

so we're looking at a circle centre (2,-1), radius 3.

now, join up all the midpoints of the the chords to the centre of the circle: these lines will be perpendicular to the chords right? (a line from the centre of a circle bisecting a chord is also perpendicular to it)

so, what you get is a 3,3,4 triangle; and we want to find the perpendicular height.

the height is root(3^2-2^2) = root 5

now, all these lengths to the midpoints are equal since they all make the same 3,3,4 triangle, but just in a different place.

ie the locus is a circle centre (2,-1), raduis root 5

ie. the equation is: (x-2)^2 +(y+1)^2= 5

 

[108796]

2. A is a point where the circle with equation x^2 + y^2 = 16 cuts the X-axis. Find the locus of the midpoints of all chords of this circle that contain the point A.

A can be (4,0) or (-4,0)

for A(4,0), the midpoint is ({x+4}/2,y/2), where x^2+y^2=16 (ie x,y lies on circle).
similar technique to a previous question:
let K=(x+4)/2, L = y/2
then K^2+L^2 = (x^2+8x+16)/4+y^2/4
= (x^2+y^2)/4+2x+4
= 2x+8 since x^2+y^2=16
now, K = (x+4)/2, so rearranging gives: x=2K-4. Plug this back in to give:

K^2+L^2 = 4K-8+8
or K^2-4K +L^2=0
or completing the square, (K-2)^2+L^2=4
i.e. it is a circle centre (2,0), radius 2!

for A(-4,0), its similar: just a reflection about the y-axis. SO;
(K+2)^2+L^2=4
ie a circle centre (-2,0), radius 2.

or if u want to go thru the algebra again:

let K=(x-4)/2, L = y/2
then K^2+L^2 = (x^2-8x+16)/4+y^2/4
= (x^2+y^2)/4-2x+4
= -2x+8 since x^2+y^2=16
now, K = (x-4)/2, so rearranging gives: x=2K+4, of -x=-4-2K. Plug this back in to give:

K^2+L^2 = -8-4K+8
or K^2+4K +L^2=0
or completing the square, (K+2)^2+L^2=4
i.e. it is a circle centre (2,0), radius 2!

 

[108796]

1. S is the point (4,0) and d is the line x = 1. The point P(x,y) moves so that the distance of P to d is half the distane of P to S. Show the equation of the locus of P is 3x^2 - y^2 = 12. (This curve is a hyperbola).

the distance from P(x,y) to d is: (x-1) [perpendicular ditance, parallel to x-axis so no y component]

the distance from P(x,y) to S is sqrt[(x-4)^2+y^2]

now, distance from P to d is half distance from P to S:

2(x-1) = sqrt[(x-4)^2+y^2]
squaring both sides and simplifying:

4x^2-8x+4 = x^2-8x+16+y^2

or: 3x^2 -y^2 = 12 QED

 

[Arithela]

thank you, I'll go through your solutions when I get more time.

 

[chewy123]

hi, can you please elaborate on this part:

the distance from P(x,y) to d is: (x-1) [perpendicular ditance, parallel to x-axis so no y component]