Question

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

• A. $16x^2+y^2+10xy=2$
• B. $16x^2-y^2-10xy=2$
• C. $16x^2-y^2+10xy=2$
• D. None of these

Answer 1

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

Let $y=4x+c$ meets $xy=1$ at two points $A$ and $B$.
$\Rightarrow A(t_1,\frac{1}{t_1}),B(t_2,\frac{1}{t_2})$
Therefore coordinates of $P$ are $(\frac{{2t}_1+t_2}{2+1},\frac{2.\frac{1}{t_1}+1.\frac{1}{t_2}}{2+1})=(h,k)$ (say)
$\therefore h=\frac{{2t}_1+t_2}{3}$ and $k=\frac{{2t}_2+t_1}{{3t}_1{t}_2}$ ...$\left(1\right)$
Also $(t_1,\frac{1}{t_1})$ and $(t_2,\frac{1}{t_2})$ lies on $y=4x+c$
$\Rightarrow \frac{\frac{1}{t_2}-\frac{1}{t_1}}{t_2-t_1}=\frac{1}{t_1{t}_2}=4\Rightarrow t_1{t}_2=-\frac{1}{4}$ ...$\left(2\right)$
From equation $\left(2\right)$
$t_1=2h+\frac{k}{4}$ and $t_2=-h-\frac{k}{2}$ ...$\left(3\right)$
From equation $\left(2\right)$ and $\left(3\right)$
$(-h-\frac{k}{2})(2h+\frac{k}{4})=-\frac{1}{4}\Rightarrow -\left(\frac{2h+k}{2}\right)\left(\frac{8h+k}{4}\right)=-\frac{1}{4}\Rightarrow (2h+k)(8h+k)=2\Rightarrow {16h}^2+k^2+10hk=2$
Hence, required locus is ${16x}^2+y^2+10xy=2$