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JEE Advanced 2015Paper 2
Exam
JEE Advanced
Year
2015
Paper
Paper 2
Subject
Mathematics
Chapter
Hyperbola

JEE Advanced 2015 · Paper 2 · Mathematics · Hyperbola

Consider the hyperbola H: x ^ 2 - y ^ 2 = 1 and a circle S with...

Consider the hyperbola H:x2−y2=1H: {x}^{2} - {y}^{2} = 1 and a circle S with center N(x2,0).N\left({x}_{2}, 0\right) . Suppose that H and S touch each other at point P(x1,y1)P\left({x}_{1}, {y}_{1}\right) with x1>1andy1>0.{x}_{1} > 1and{y}_{1} > 0 . The common tangent to H and S at P intersects the x - axis at point M. If (l,m)( l, m ) is the centroid of the triangle δPMN,\delta PMN, then the correct expression(s) is(are)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 eachConsider 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

Options

  1. Adldx1=1−13x12forx1>1\dfrac{dl}{d{x}_{1}} = 1 - \dfrac{1}{3{x}_{1}^{2}}for{x}_{1} > 1
  2. Bdmdx1=x13(x12−1)forx1>1\dfrac{dm}{d{x}_{1}} = \dfrac{{x}_{1}}{3\left(\sqrt{{x}_{1}^{2} - 1}\right)}for{x}_{1} > 1
  3. Cdldx1=1+13x12forx1>1\dfrac{dl}{d{x}_{1}} = 1 + \dfrac{1}{3{x}_{1}^{2}}for{x}_{1} > 1
  4. Ddmdy1=13fory1>0\dfrac{dm}{d{y}_{1}} = \dfrac{1}{3}for{y}_{1} > 0

Correct answer

A. dldx1=1−13x12forx1>1\dfrac{dl}{d{x}_{1}} = 1 - \dfrac{1}{3{x}_{1}^{2}}for{x}_{1} > 1

Step-by-step solution


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
Equation of tangent at P on hyperbola.
xx1−yy1=1x{x}_{1} - y{y}_{1} = 1
Point M(1x1,0)M\left(\dfrac{1}{{x}_{1}}, 0\right)
Equation of normal at P
xx1+yy1=2\dfrac{x}{{x}_{1}} + \dfrac{y}{{y}_{1}} = 2
Since (x2,0)( {x}_{2}, 0 ) satisfies its
x2=2x1{x}_{2} = 2{x}_{1}
Centroid \left(l, m\right) ≡ (x1+13x1,y13)\left({x}_{1} + \dfrac{1}{3{x}_{1}}, \dfrac{{y}_{1}}{3}\right)
dldx1=1−13x12\dfrac{dl}{d{x}_{1}} = 1 - \dfrac{1}{{3x}_{1}^{2}}
dmdy1=13\dfrac{dm}{d{y}_{1}} = \dfrac{1}{3}
x12−y12=1{x}_{1}^{2} - {y}_{1}^{2} = 1
y1=x12−1{y}_{1} = \sqrt{{x}_{1}^{2} - 1}
m=x12−13m = \dfrac{\sqrt{{x}_{1}^{2} - 1}}{3}
dmdx1=x13x12−1\dfrac{dm}{d{x}_{1}} = \dfrac{{x}_{1}}{3\sqrt{{x}_{1}^{2} - 1}}
Alternative Method

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
l=3secθ+cosθ3l = \dfrac{3sec \theta + cos \theta}{3}
m=tanθ3=y13m = \dfrac{tan \theta}{3} = \dfrac{{y}_{1}}{3}
dldx1=3secθtanθ−sinθ3secθtanθ=1−13sec⁡2θ=1−13x12\dfrac{dl}{d{x}_{1}} = \dfrac{3sec \theta tan \theta - sin \theta}{3sec \theta tan \theta} = 1 - \dfrac{1}{3\sec^{2}\theta} = 1 - \dfrac{1}{{3x}_{1}^{2}}
dmdx1=sec⁡2θ3secθtanθ=cosecθ3=x13x12−1\dfrac{dm}{d{x}_{1}} = \dfrac{\dfrac{\sec^{2} \theta}{3}}{sec \theta tan \theta} = \dfrac{cosec\theta}{3} = \dfrac{{x}_{1}}{3\sqrt{{x}_{1}^{2} - 1}}
dmdy1=13\dfrac{dm}{d{y}_{1}} = \dfrac{1}{3}
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