JEE Advance - Mathematics (2015 - Paper 2 Offline - No. 10)

Consider the hyperbola $$H:{x^2} - {y^2} = 1$$ and a circle $$S$$ with center $$N\left( {{x_2},0} \right)$$. Suppose that $$H$$ and $$S$$ touch each other at a point $$P\left( {{x_1},{y_1}} \right)$$ with $${{x_1} > 1}$$ and $${{y_1} > 0}$$. The common tangent to $$H$$ and $$S$$ at $$P$$ intersects the $$x$$-axis at point $$M$$. If $$(l, m)$$ is the centroid of the triangle $$PMN$$, then the correct expressions(s) is(are)
$${{dl} \over {d{x_1}}} = 1 - {1 \over {3x_1^2}}$$ for $${x_1} > 1$$
$${{dm} \over {d{x_1}}} = {{{x_1}} \over {3\left( {\sqrt {x_1^2 - 1} } \right)}}$$ for $${x_1} > 1$$
$${{dl} \over {d{x_1}}} = 1 + {1 \over {3x_1^2}}$$ for $${x_1} > 1$$
$${{dm} \over {d{y_1}}} = {1 \over 3}$$ for $${y_1} > 0$$

Explanation

JEE Advanced 2015 Paper 2 Offline Mathematics - Hyperbola Question 14 English Explanation

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

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