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. $16x^2+y^2+10xy+2=0$
  
• B. $16x^2-y^2+10xy=2$
  
• C. $16x^2-y^2-10xy=2$
  
• D. $16x^2+y^2+10xy=2$
  

Answer 1

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


Let $y=4x+c$ meets $xy=1$ at two points A and B.Let the points on the hyperbola be $A(t_1,\frac{1}{t_1}),B(t_2,\frac{1}{t_2})$

P divide the line segment $BA$ in ratio $2:1$

$\therefore$ Cooridnates of P are
$(\frac{2t_1+t_2}{2+1},\frac{2\cdot \frac{1}{t_1}+1\cdot \frac{1}{t_2}}{2+1})\equiv (h,k)\text{(say})$

$\Rightarrow h=\frac{2t_1+t_2}{3}\text{and}k=\frac{2t_2+t_1}{3t_1{t}_2}\cdot \cdot \cdot \left(1\right)$

Points $A(t_1,\frac{1}{t_1}),B(t_2,\frac{1}{t_2})$ lie on the line $y=4x+c$

$\Rightarrow \text{Slope}=\frac{\frac{1}{t_2}-\frac{1}{t_1}}{t_2-t_1}=4$
$\Rightarrow \frac{1}{t_1{t}_2}=-4\Rightarrow t_1{t}_2=\frac{-1}{4}\cdot \cdot \cdot \left(2\right)$

From Eq. (1) and Eq.(2),
$\Rightarrow t_1=2h+\frac{k}{4}\text{and}{t}_2=-h-\frac{k}{2}\cdot \cdot \cdot \left(3\right)$

From Eq. (2) and (3)
$\Rightarrow (-h-\frac{k}{2})(2h+\frac{k}{4})=\frac{-1}{4}$

$\Rightarrow (\frac{2h+k}{2})(\frac{8h+k}{4})=\frac{-1}{4}$

$\Rightarrow (2h+k)(8h+k)=2$

$\Rightarrow 16h^2+k^2+10hk=2$

Hence, required locus is $16x^2+y^2+10xy=2$