QJEEAdv15Ma7731: JEE Advanced 2015 Mathematics - Class 11 JEEAdv15Ma7731 Solution: Consider the hyperbola and a circle with center . Suppose that…

Equation of family of circles touching hyperbola at (x₁, y₁) is

(x $-$ x₁)² + (y $-$ y₁)² + $\lambda$ (xx₁ $-$ yy₁ $-$ 1) =0

Now, its centre is (x₂, 0).

$\therefore$ $\left[ {{{ - (\lambda {x_1} - 2{x_1})} \over 2},{{ - ( - 2{y_1} - \lambda {y_1})} \over 2}} \right] = ({x_2},0)$

$\Rightarrow 2{y_1} + \lambda {y_1} = 0 \Rightarrow \lambda = - 2$

and $2{x_1} - \lambda x = 2{x_2} \Rightarrow {x_2} = 2{x_1}$

$\therefore$ $P({x_1},\sqrt {x_1^2 - 1} )$

and $N({x_2},0) = (2{x_1},0)$

As tangent intersect X-axis at $M\left( {{1 \over x},0} \right)$ .

Centroid of $\Delta PWN = (l,m)$

$\Rightarrow \left( {{{3{x_1} + {1 \over {{x_1}}}} \over 3},{{{y_1} + 0 + 0} \over 3}} \right) = (l,m)$

$\Rightarrow l = {{3{x_1} + {1 \over {{x_1}}}} \over 3}$

On differentiating w.r.t. x₁, we get

${{dl} \over {d{x_1}}} = {{3 - {1 \over {x_1^2}}} \over 3}$

$\Rightarrow {{dl} \over {d{x_1}}} = 1 - {1 \over {3x_2^1}}$ , for x₁ > 1

and $m = {{\sqrt {x_1^2 - 1} } \over 3}$

On differentiating w.r.t. x₁, we get

${{dm} \over {d{x_1}}} = {{2{x_1}} \over {2 \times 3\sqrt {x_1^2 - 1} }} = {{{x_1}} \over {3\sqrt {x_1^2 - 1} }}$ , for x₁ > 1

Also, $m = {{{y_1}} \over 3}$

On differentiating w.r.t. y₁, we get

${{dm} \over {d{y_1}}} = {1 \over 3}$ , for y₁ > 0