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)

$16x^{2} + 10xy + y^{2} = 2$

Answer 2

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

Suppose this meets $xy = 1$ ......(ii) in $A(x_{1},y_{1})$ and

$B(x_{2},y_{2})$

Eliminating y between (i) and (ii), we get $\frac{1}{x} - k = 4(x - h)$

⇒ $1 - xk = 4x^{2} - 4hx$ ⇒ $4x^{2} - (4h - k)x - 1 = 0$ ......(iii)

This has two roots say $x_{1},x_{2}$; $x_{1} + x_{2} = \frac{4h - k}{4}$ ......(iv) and $x_{1}x_{2} = - \frac{1}{4}$ ......(v)

Also, $\frac{2x_{1} + x_{2}}{3} = h$ [$\because$ P divides AB in the ratio 1 : 2]

i.e., $2x_{1} + x_{2} = 3h$ ......(vi)

(vi) – (iv) gives, $x_{1} = 3h - \frac{4h - k}{4} = \frac{8h + k}{4}$ and

$x_{2} = 3h - 2.\frac{8h + k}{4} = - \frac{2h + k}{2}$Putting in (v), we get

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

⇒ $(8h + k)(2h + k) = 2$ ⇒ $16h^{2} + 10hk + k^{2} = 2$

$\therefore$ Required locus of $P(h,k)$ is $16x^{2} + 10xy + y^{2} = 2$