Question 1019640
We use the following form of the straight line:  {{{x/a + y/b = 1}}}, where a is the value of the x-intercept and b is the value of the y-intercept.

The equation of the straight line passing through the point (2,4) is given by {{{2/a + 4/b = 1}}} as per the previous statement.

Solving for b in terms of a, 
{{{4/b = 1-2/a}}},
==> {{{4/b = (a-2)/a}}}
==> {{{b/4 = a/(a-2)}}}
==> {{{b = (4a)/(a-2)}}}. (Equation A)
This gives the relationship between the x- and y-intercepts  of the line.

Now the midpoint of the x-intercept (a,0) and the y-intercept (0,b) is the point (a/2, b/2).
Substituting Equation A into the y-coordinate of the point (a/2, b/2), we get
({{{a/2}}}, {{{(2a)/(a-2)}}})
Now let x = a/2 and y = {{{(2a)/(a-2)}}}
==> a = 2x
==> y = {{{(2(2x))/(2x-2)}}}
==> y = {{{(2x)/(x-1)}}}
This is the equation of the locus satisfying the given conditions.
This is a hyperbola with horizontal asymptote of y = 2, and vertical asymptote of x = 1.