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JEE Main 201510 AprilOnline
Exam
JEE Main
Year
2015
Date
10 April
Mode
Online
Subject
Mathematics
Chapter
Parabola

JEE Main 2015 · 10 April · Online · Mathematics · Parabola

If the tangent to the conic, y - 6 = x ^ 2 at ( 2, 10 ) touches the...

If the tangent to the conic, y−6=x2y - 6 = {x}^{2} at (2,10)( 2, 10 ) touches the circle, x2+y2+8x−2y=k{x}^{2} + {y}^{2} + 8x - 2y = k (for some fixed kk) at a point (α,β);\left(\alpha, \beta\right); then (α,β)( \alpha, \beta ) isIf the tangent to the conic, y - 6 = x ^ 2 at ( 2, 10 ) touches the circle, x ^ 2 + y ^ 2 + 8x - 2y = k (for some fixed

Options

  1. A(−717,617)\left(- \dfrac{7}{17}, \dfrac{6}{17}\right)
  2. B(−817,217)\left(- \dfrac{8}{17}, \dfrac{2}{17}\right)
  3. C(−617,1017)\left(- \dfrac{6}{17}, \dfrac{10}{17}\right)
  4. D(−417,117)\left(- \dfrac{4}{17}, \dfrac{1}{17}\right)

Correct answer

B. (−817,217)\left(- \dfrac{8}{17}, \dfrac{2}{17}\right)

Step-by-step solution

The equation of tangent (T=0T = 0) would be 12(y+10)−6=2x\dfrac{1}{2}\left(y + 10\right) - 6 = 2x

4x−y+2=04x - y + 2 = 0

If the tangent to the conic, y - 6 = x ^ 2 at ( 2, 10 ) touches the circle, x ^ 2 + y ^ 2 + 8x - 2y = k (for some fixed

The center of the circle is (−4,1)( - 4, 1 ) and the point of contact would be the foot of perpendicular from (−4,1)( - 4, 1 ) on 4x−y+2=04x - y + 2 = 0

α+44=β−1−1=−(−16−1+242+12)\dfrac{\alpha + 4}{4} = \dfrac{\beta - 1}{- 1} = - \left(\dfrac{- 16 - 1 + 2}{{4}^{2} + {1}^{2}}\right)

α+44=1517\dfrac{\alpha + 4}{4} = \dfrac{15}{17} and β−1−1=1517\dfrac{\beta - 1}{- 1} = \dfrac{15}{17}

α=−817\alpha = - \dfrac{8}{17}, β=−1517+1=217\beta = \dfrac{- 15}{17} + 1 = \dfrac{2}{17}

Hence option (−817,217)\left(- \dfrac{8}{17}, \dfrac{2}{17}\right) is correct.
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