A variable line, drawn through the point of intersection of the straight lines (x/a)+(y/b) = 1 and (x/b)+(y/a)=1, meets the coordinate axes in A & B . We have to Show that the locus of the mid point of AB is the curve 2xy(a + b) = ab(x + y)
I tried as
Equation of line can be written as L$_1$+kL$_2$= 0
From this i got A(a,0) and B(0,b).
Best Answer
You have proceeded rightly. The equation of the variables line $AB $ is $(bx+ay-ab) + k (ax+by-ab) =0$
Then we get $A (\frac {ab (1+k)}{b + ak},0)$ and $B (0,\frac {ab (1+k)}{a +bk}) $. Let $(h,k) $ be the mid point of $AB $.
Thus $h =\frac {ab (1+k)}{2 (b+ak)} $ and $k=\frac {ab (1+k)}{2 (a+bk)} $. Now, $$\frac{1}{2}[\frac {1}{h}+\frac {1}{k}] = \frac {1}{2}[\frac {2 (b+ak)+2 (a+bk)}{ab (1+k)}] = \frac {a+b}{ab} $$ and thus the locus of the midpoint is $$\boxed {2xy(a + b) = ab(x + y)} $$ Hope it helps .