y² = 2x

y = √2x or y = -√2x

So, any point on the curve has coordinates (x,√2x) or (x, -√2x), depending on which side of the parabola it is on.

Given A(2,6) ,

Midpoint of AB has 2 possibilities :


( (2+x)/2, (6 + √2x)/2 ) or ( (2+x)/2, (6 - √2x)/2)

Which simplifies to :

( (1+½x, 3 + √x/√2 ) or ( (1+½x , 3 - √x/√2)


So any point on the new curve that's traced out has either of the above coordinates.

Now, how do we express the y-coordinate in terms of the x-coordinate?

( (1+½x, 3 + √x/√2 ) or ( (1+½x , 3 - √x/√2)


The idea here is to subtract the constant from each y-coordinate first,then square it, the reverse of what we did at first.

We work towards obtaining the x coordinate on the right hand side.

i.e

Let Y be the y-coordinate, X be the x-coordinate,


Y = 3 + √x/√2 or Y = 3 - √x/√2

Y - 3 = √x/√2 or Y - 3 = -√x/2

(Y - 3)² = (√x/√2)² or (Y - 3)² = (-√x/√2)²

(Y - 3)² = x/2 or (Y - 3)² = x/2

1 + (Y - 3)² = 1 + ½x

1 + (Y - 3)² = X

(Y - 3)² = X - 1



Therefore, the equation of the curve traced by the midpoint of AB is that of a parabola,


(y - 3)² = x - 1

What we also realise is that the origin (0,0) lies on the curve and is its vertex.

i.e when x = 0, y² = 0
y = 0

The midpoint of A(2,6) and the (0,0) would give us a point which is the vertex of the new parabola we would obtain.


Midpoint of A(2,6) and (0,0)

= ( (2 + 0)/2, (6 + 0)/2 )

= (1,3)

Using the equation of parabola of the form (y - k)² + h = x , where (h,k) are the coordinates of the vertex,


We get (y - 3)² + 1 = x

(y - 3)² = x - 1


The same result as the first method.

Thank you so much you explained it well! I understand now.

Welcome!

Another way to think of it is :


When we plot all the midpoints of AB (i.e between A and every possible point on the curve), we are just finding all the averages of the y and x coordinates of A and B.


As you move from the origin along the curve rightwards, the y-coordinates will increase/decrease as the x-coordinate increases.


This means that the midpoints' y-coordinates will increase/decrease and the x-coordinates will increase in the same fashion.


This results in the tracing of a parabola shape similar to the original cirve.

( (1+½x, 3 + √x/√2 ) or ( (1+½x , 3 - √x/√2)

This means that for the x-coordinate, we are halving its value, then adding 1.

For the y-coordinate, we are having its value, then adding 3.


Halving the values would result in the same parabola shape traced out since they are all scaled down in the same way.

Eg.

Coordinates (0,0) , (1,√2) , (2,2), (3,√6), (4,2√2) just becomes (0,0) , (½, ½√2) , (1,1) (1½, ½√6), (2, √2)

We get the same parabola shape. It's equivalent to compressing the original curve.


Adding 1 or 3 would also result in the same shape traced out since we are just shifting/translating all the coordinates 1 unit to the right, and the y-coordinates 3 units upwards.

i.e the parabola gets shifted diagonally upwards to the right