Polynomials (1 Viewer)

[jaychouf4n]

Can someone help me solve

Find the locus of the midpoint of parallel chords to the curve xy=c^2. You may assume the equation of the chord is y=mx+b, where m is a constant.

In the back they solve simultaneously y=mx+b and xy=c^2

Why??? They don't meet >< do they?

omg wtfbbq

 

[jet]

Ok so the reason for that is because they are chords of the hyperbola. So while they intersect, the chords, which are parallel do not.
You will find the midpoint of each chord and then this should give you a locus (a hyperbola i think from memory)
I'll tryand do it for you now

 

[jet]

Its actually a straight line.
So we have xy=c^2 and y=mx+b
Solving these two we have
x(mx + b) = c^2
mx^2 + bx-c^2=0
Now, the two roots of this equation are the x-values of the points of intersection of the hypercbola and the chord.
Hence the sumof roots/2 will give the x value, X say, of the midpoint.
Sumof roots/2 = -b/(2m)
Hence X= -b/(2m)
Now, the Y value of the midpoint is:
**Y=mX + b
=m(-b/(2m))+ b
=b/2

But X=-b/2m
Hence b=-2mX

Therefore Y=-mX is what i got. This seems right to me. Do you have the answer?

EDIT
**or you might say
b=2mX
but Y=mX + b
Y=mX-2mX
=-mX, which gives the answer a tad faster