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JEE Main 202225 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
25 June
Shift
Morning Shift
Subject
Mathematics
Chapter
Parabola

JEE Main 2022 · 25 June · Morning Shift · Mathematics · Parabola

Let x = 2t, y = t ^ 2 3 be a conic. Let S be the focus and B be the...

Let x=2t,y=t23x = 2t, y = \dfrac{{t}^{2}}{3} be a conic. Let SS be the focus and BB be the point on the axis of the conic such that SA⊥BASA ⊥ BA, where AA is any point on the conic. If kk is the ordinate of the centroid of the δSAB\delta SAB, then lim⁡t→1k\lim_{t \to 1}k is equal toLet x = 2t, y = t ^ 2 3 be a conic. Let S be the focus and B be the point on the axis of the conic such that SA ⊥ BA ,

Options

  1. A1718\dfrac{17}{18}
  2. B1918\dfrac{19}{18}
  3. C1118\dfrac{11}{18}
  4. D1318\dfrac{13}{18}

Correct answer

D. 1318\dfrac{13}{18}

Step-by-step solution

The parametric form x=2t,y=t23x = 2t, y = \dfrac{{t}^{2}}{3} represents the parabola x2=12y{x}^{2} = 12y

Let x = 2t, y = t ^ 2 3 be a conic. Let S be the focus and B be the point on the axis of the conic such that SA ⊥ BA ,

Given AS⊥ABAS ⊥ AB

So, mAS⋅mAB=−1{m}_{AS} \cdot {m}_{AB} = - 1

⇒(3−t23)(0−2t)⋅(α−t23)(0−2t)=−1⇒ \dfrac{\left(3 - \dfrac{{t}^{2}}{3}\right)}{\left(0 - 2t\right)} \cdot \dfrac{\left(\alpha - \dfrac{{t}^{2}}{3}\right)}{\left(0 - 2t\right)} = - 1

⇒3α=27t2+t4t2−9⇒ 3\alpha = \dfrac{27{t}^{2} + {t}^{4}}{{t}^{2} - 9}

Ordinate of centroid of δSAB=k=α+t23+33=9+3α+t29\delta SAB = k = \dfrac{\alpha + \dfrac{{t}^{2}}{3} + 3}{3} = \dfrac{9 + 3\alpha + {t}^{2}}{9}

lim⁡t→1k=lim⁡t→119(9+t2+27t2+t4(t2−9))=1318\lim_{t \to 1}k = \lim_{t \to 1}\dfrac{1}{9}\left(9 + {t}^{2} + \dfrac{27{t}^{2} + {t}^{4}}{\left({t}^{2} - 9\right)}\right) = \dfrac{13}{18}

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