Question

How do you find the locus of the center of the hyperbola having asymptotes given by `y-x tan alpha+1=0 and y-x tan(alpha+pi/4)+2=0`, where `alpha` varies?

Answer 1

The equations of the locus are:

`x=cosalpha(cosalpha-sinalpha)`

`y=-1+sinalpha(cosalpha-sinalpha)`

Explanation:

By definition the center of the hyperbola is the point where the asymptotes cross, so that they solve the system:

`y-xtanalpha+1 = 0`
`y-xtan(alpha+pi/4)+2=0`

`y-xtanalpha = -1`
`y-xtan(alpha+pi/4)=-2`

Subtracting member by member:

`x(tan(alpha+pi/4)-tanalpha)= 1`

So:

`x=1/(tan(alpha+pi/4)-tanalpha)`

Now note that:

`1/(tan(alpha+pi/4)-tanalpha)= 1/(sin(alpha+pi/4)/cos(alpha+pi/4)-sinalpha/cosalpha)= (cos(alpha+pi/4)cosalpha)/(sin(alpha+pi/4)cosalpha-sinalphacos(alpha+pi/4))=cosalpha(cosalphacos(pi/4)-sinalphasin(pi/4))/sin(alpha+pi/4-alpha)=cosalpha(cosalpha-sinalpha)`

so:

`x=cosalpha(cosalpha-sinalpha)`

`y=-1+tanalphacosalpha(cosalpha-sinalpha)=-1+sinalpha(cosalpha-sinalpha)`

Are the equations of the locus.

Answer 2

This locus is the circle, with center at `(1/2, -3/2)` and
radius `1/sqrt2`.

Explanation:

The asymptotes meet at the center of the hyperbola.

Let `t = tan alpha`. Use `tan(A+B)=(tan A + tan B)/(1-tan A tan B)` and `tan(pi/4) = 1`.

The coordinates (x, y) of the center are given by

`y= xt-1=x(t+1)/(1-t.)-2`. Eliminating t,

`y=x((1+(y+1)/x)/((1-(y+1)/x)))-2`

`=x((x+y+1)/(x-y-1))-2`. Cross multiplying,

`xy-y^2-y=x^2+xy+x-2(x-y-1)`. Simplifying,

`x^2+y^2-x+3y+2=0`. In the standard form,

`(x-1/2)^2+(y+3/2)^2=(1/sqrt2)^2`

This locus represents the circle with center at `(1/2, -3/2)` and

radius `1/sqrt2`.