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 ratio $1:2$ is

• A. $16x^2+10xy+y^2=2$
• B. $16x^2-10xy+y^2=2$
• C. $16x^2+10xy+y^2=4$
• D. none of these

Answer 1

Answer: (A)

Let $P(h,k)$ be any point on the locus. Equation of the line through $P$ and having slope $4$ is

$y-k=4(x-h)...$(1)
Suppose, this line meets $xy=\mathrm{1...}$(2)
in $A\left(x_1,y_1\right)$ and $B\left(x_2,y_2\right)$
Eliminating $y$ from (1) and (2), we get
$\frac{1}{x}-k=4(x-h)$
$\Rightarrow 4x^2-(4h-k)x-1=\mathrm{0...}$(3)
Since $x_1$ and $x_2$ are the roots of $(3)$
$\therefore x_1+x_2=\frac{4h-k}{4}....$(4)
and, $x_1{x}_2=\frac{-1}{4}....$(5)
Also, $\left(\frac{8h+k}{2}\right)\left(-\frac{2h+k}{2}\right)=-\frac{1}{4}=h$
or $2x_1+x_2=3h....$ (6)
From (4) and (6), we get
$x_1=3h-\frac{(4h-k)}{4}=\frac{8h+k}{4}$
and, $x_2=3h-\frac{(8h+k)}{2}=\frac{-(2h+k)}{2}$
Putting these values in (5), we get
$\left(\frac{8h+k}{2}\right)\left(-\frac{2h+k}{2}\right)=-\frac{1}{4}$
$\Rightarrow (8h+k)(2h+k)=2$ or $16h^2+10hK+k^2=2$
Thus, equation of required locus is
$16x^2+10xy+y^2=2$.