Question

A variable straight line of slope $4$ intersects the hyperbola $xy=1$ at two points. The locus of the point which divides the line segment between these two points in the $1:2$ is

• A. $8x^2-y^2+2xy=2$
• B. $16x^2+y^2+10xy=2$
• C. $8x^2+y^2+2xy=2$
• D. $16x^2-y^2+10xy=2$

Answer 1

The correct option is B $16x^2+y^2+10xy=2$

Let two points on the hyperbola $A\equiv (t_1,\frac{1}{t_1})$ and $B\equiv (t_2,\frac{1}{t_2})$
$\therefore$ slope of $AB=4$
$\Rightarrow t_1{t}_2=-\frac{1}{4}\cdots \left(1\right)$
Now let $P(h,k)$ on the line $AB$

Case $1:AP:PB=1:2$
$h=\frac{2t_1+t_2}{3},k=\frac{t_1+2t_2}{3t_1{t}_2}\Rightarrow t_1=\frac{8h+k}{4},t_2=-\frac{(k+2h)}{2}$
using $\left(1\right)$
$(8h+k)(k+2h)=2$
Hence locus will be
$16x^2+y^2+10xy=2$

Case $2:AP:PB=2:1$
$h=\frac{t_1+2t_2}{3},k=\frac{2t_1+t_2}{3t_1{t}_2}\Rightarrow t_1=-\frac{(2h+k)}{2},t_2=\frac{8h+k}{4}$
using $\left(1\right)$
$(8h+k)(k+2h)=2$
Hence locus will be
$16x^2+y^2+10xy=2$