Theoretical Coordinate Geometry

Jan 2009
10
1
Okay, I came across this question in an old mock paper. The problem is that I couldn't do it and neither could my teacher. Any help would be appreciated.

The line y=2kx cuts the curve \(\displaystyle y=x^2-4x+1\) at the points a and b. Without finding the co-ordinates of a and b, find, in terms of k, the midpoint of [ab].

Now I tried putting \(\displaystyle 2kx=x^2-4x+1\)
and this worked down to \(\displaystyle x=2+k \pm \sqrt[2]{k^2+4k+3}\)
But I feel that would be finding the co-ordinates of a and b.
Again, any help would be appreciated.
 
Oct 2009
769
87
Opinion

Without knowing what a and b are, how can you determine what the midpoint is?
 
Jan 2009
10
1
That's what I was wondering. Maybe there's some relationship between the midpoint and the axis of the curve...
 

masters

MHF Helper
Jan 2008
2,550
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Big Stone Gap, Virginia
Okay, I came across this question in an old mock paper. The problem is that I couldn't do it and neither could my teacher. Any help would be appreciated.

The line y=2kx cuts the curve \(\displaystyle y=x^2-4x+1\) at the points a and b. Without finding the co-ordinates of a and b, find, in terms of k, the midpoint of [ab].

Now I tried putting \(\displaystyle 2kx=x^2-4x+1\)
and this worked down to \(\displaystyle x=2+k \pm \sqrt[2]{k^2+4k+3}\)
But I feel that would be finding the co-ordinates of a and b.
Again, any help would be appreciated.
Hi FreeT,

It would seem that you are on the right track.

Once you have found the coordinates of a and b in terms of k,
you can then determine the midpoint of ab.

\(\displaystyle a(x_1, y_1)\) and \(\displaystyle b(x_2, y_2)\)

Midpoint of ab = \(\displaystyle \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)
 

Plato

MHF Helper
Aug 2006
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Solve the equation \(\displaystyle x^2-4x+1=2kx\) for \(\displaystyle x\) in terms of \(\displaystyle k\).
That will give the \(\displaystyle x\)-coordinates of \(\displaystyle a~\&~b\) in terms of \(\displaystyle k\).
Use the equation \(\displaystyle y=2kx\) to find the \(\displaystyle y\)-coordinates of \(\displaystyle a~\&~b\) in terms of \(\displaystyle k\).
The midpoints are just the averages.
 

awkward

MHF Hall of Honor
Mar 2008
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Okay, I came across this question in an old mock paper. The problem is that I couldn't do it and neither could my teacher. Any help would be appreciated.

The line y=2kx cuts the curve \(\displaystyle y=x^2-4x+1\) at the points a and b. Without finding the co-ordinates of a and b, find, in terms of k, the midpoint of [ab].

Now I tried putting \(\displaystyle 2kx=x^2-4x+1\)
and this worked down to \(\displaystyle x=2+k \pm \sqrt[2]{k^2+4k+3}\)
But I feel that would be finding the co-ordinates of a and b.
Again, any help would be appreciated.
The tricky point is avoiding actually solving the equation. Here is one way.

By substitution,
\(\displaystyle 2kx = x^2 - 4x +1\)
so
\(\displaystyle x^2 - (4+2k)x + 1 = 0\)

If a and b are the roots of this equation, then their sum must be the negative of the coefficient of x, i.e.
\(\displaystyle a+b = 4+2k\).

From here you can find \(\displaystyle \frac{a+b}{2}\) and the associated value of y.
 
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