Consider the hyperbola H : x 2 - y 2 = 1 and a circle S with center N ( x 2 , 0 ) . Suppose that H and S touch each other at a point P ( x 1 , y 1 ) with x 1 1 and y 1 0 . The common tangent to H and S at P intersects the x -axis at p

Question

Consider the hyperbola $H : x^{2} - y^{2} = 1$ and a circle $S$ with center $N (x_{2}, 0)$ . Suppose that $H$ and $S$ touch each other at a point $P (x_{1}, y_{1})$ 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 expression(s) is/are [2015]

Smart Achievers · Accepted Answer

(1, 2, 4)

$H : x^{2} - y^{2} = 1$ is a hyperbola and $S : Circle with centre N (x_{2}, 0) .$ Common tangent to $H$ and $S$ at $P (x_{1}, y_{1})$ is

$x x_{1} - y y_{1} = 1$ $\Rightarrow m_{1} = \frac{x_{1}}{y_{1}}$

Now, radius of circle $S$ with centre $N (x_{2}, 0)$ through the point of contact $(x_{1}, y_{1})$ is perpendicular to the tangent.

$∴$ $m_{1} m_{2} = - 1$ $\Rightarrow \frac{x_{1}}{y_{1}} \times \frac{0 - y_{1}}{x_{2} - x_{1}} = - 1$

$\Rightarrow x_{2} = 2 x_{1}$

$∵$ $M$ is the point of intersection of tangent at $P$ and $x$ -axis.

$∴$ $M (\frac{1}{x_{1}}, 0)$ $∵$ $Centroid of ∆ P M N is (ℓ, m)$

$∴$ $x_{1} + \frac{1}{x_{1}} + x_{2} = 3 ℓ$ and $y_{1} = 3 m$

$\Rightarrow \frac{1}{3} (3 x_{1} + \frac{1}{x_{1}}) = ℓ$ and $\frac{y_{1}}{3} = m [∵ x_{2} = 2 x_{1}]$

$∴$ $\frac{d ℓ}{d x_{1}} = 1 - \frac{1}{3 x_{1}^{2}}, \frac{d m}{d y_{1}} = \frac{1}{3}$

Also $(x_{1}, y_{1})$ lies on $H$ , $∴$ $x_{1}^{2} - y_{1}^{2} = 1$ $\Rightarrow y_{1} = \sqrt{x_{1}^{2} - 1}$

$∴$ $m = \frac{1}{3} \sqrt{x_{1}^{2} - 1}$ $∴$ $\frac{d m}{d x_{1}} = \frac{x_{1}}{3 \sqrt{x_{1}^{2} - 1}}$

Solution

1 $\frac{d l}{d x_{1}} = 1 - \frac{1}{3 x_{1}^{2}} for x_{1} > 1$

2 $\frac{d m}{d x_{1}} = \frac{x_{1}}{3 (\sqrt{x_{1}^{2} - 1})} for x_{1} > 1$

3 $\frac{d l}{d x_{1}} = 1 + \frac{1}{3 x_{1}^{2}} for x_{1} > 1$

4 $\frac{d m}{d y_{1}} = \frac{1}{3} for y_{1} > 0$

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Solution

1 dldx1=1-13x12 for x1>1

2 dmdx1=x13(x12-1) for x1>1

3 dldx1=1+13x12 for x1>1

4 dmdy1=13 for y1>0

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1 $\frac{d l}{d x_{1}} = 1 - \frac{1}{3 x_{1}^{2}} for x_{1} > 1$

2 $\frac{d m}{d x_{1}} = \frac{x_{1}}{3 (\sqrt{x_{1}^{2} - 1})} for x_{1} > 1$

3 $\frac{d l}{d x_{1}} = 1 + \frac{1}{3 x_{1}^{2}} for x_{1} > 1$

4 $\frac{d m}{d y_{1}} = \frac{1}{3} for y_{1} > 0$