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JEE Advanced 2017Paper 1
Exam
JEE Advanced
Year
2017
Paper
Paper 1
Subject
Mathematics
Chapter
Ellipse

JEE Advanced 2017 · Paper 1 · Mathematics · Ellipse

Columns 1, 2 and 3 contain conics, equations of tangents to the...

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.
Column 1 Column 2 Column 3
(I) x2+y2=a2{x}^{2} + {y}^{2} = {a}^{2} (i) my=m2x+amy = {m}^{2}x + a (P) \left(\dfrac{a}{{m}^{2}}, \dfrac{2a}{m}\right)</td></tr><tr><td>(II)</td> </tr> <tr> <td>(II){x}^{2} + {a}^{2}{y}^{2} = {a}^{2}</td><td>(ii)</td> <td>(ii)y = mx + a\sqrt{{m}^{2} + 1}</td><td>(Q)(−mam2+1,am2+1)</td> <td>(Q) \left(\dfrac{- ma}{\sqrt{{m}^{2} + 1}}, \dfrac{a}{\sqrt{{m}^{2} + 1}}\right)
(III) y2=4ax{y}^{2} = 4ax (iii) y=mx+a2m2−1y = mx + \sqrt{{a}^{2}{m}^{2} - 1} (R) \left(\dfrac{- {a}^{2}m}{\sqrt{{a}^{2}{m}^{2} + 1}}, \dfrac{1}{\sqrt{{a}^{2}{m}^{2} + 1}}\right)</td></tr><tr><td>(IV)</td> </tr> <tr> <td>(IV){x}^{2} - {a}^{2}{y}^{2} = {a}^{2}</td><td>(iv)</td> <td>(iv)y = mx + \sqrt{{a}^{2}{m}^{2} + 1}</td><td>(S)(−a2ma2m2−1,−1a2m2−1)</td> <td>(S) \left(\dfrac{- {a}^{2}m}{\sqrt{{a}^{2}{m}^{2} - 1}}, \dfrac{- 1}{\sqrt{{a}^{2}{m}^{2} - 1}}\right)
The tangent to a suitable conic (Column 1) at \left(\sqrt{3}, \dfrac{1}{2}\right)isfoundtobeis found to be\sqrt{3}x + 2y = 4 , then which of the following options is the only Correct combination?

Options

  1. A(IV) (iii) (S)
  2. B(II) (iv) (R)
  3. C(IV) (iv) (S)
  4. D(II) (iii) (R)

Correct answer

B. (II) (iv) (R)

Step-by-step solution

P(3,12);\left(\sqrt{3}, \dfrac{1}{2}\right); tangent 3x+2y=4\sqrt{3}x + 2y = 4

⇒(3)x+4(12)y=4⇒ \left(\sqrt{3}\right)x + 4\left(\dfrac{1}{2}\right)y = 4, comparing with (II)

⇒a=2⇒ a = 2

∴y=mx+a2m2+1∴ y = mx + \sqrt{{a}^{2}{m}^{2} + 1} is tangent for m=−32m = - \dfrac{\sqrt{3}}{2} i.e. (ii)

∴∴ Point of contact for a=2,m=−32a = 2, m = - \dfrac{\sqrt{3}}{2} is R

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