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Parabola question

2022 · 25 Jun · Shift 1 · Q37
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  5. /2022 · 25 Jun · Shift 1 · Q37

Parabola question

2022 · 25 Jun · Shift 1 · Q37

JEE MainMathematicsParabolaMCQ+4 / −1
Let x=2tx = 2tx=2t, y=t23y = {{{t^2}} \over 3}y=3t2​ be a conic. Let S be the focus and B be the point on the axis of the conic such that SA⊥BASA \bot BASA⊥BA, where A is any point on the conic. If k is the ordinate of the centroid of the Δ\DeltaΔ SAB, then lim⁡t→1k\mathop {\lim }\limits_{t \to 1} kt→1lim​k is equal to :
  1. A
    1718{{17} \over {18}}1817​
  2. B
    1918{{19} \over {18}}1819​
  3. C
    1118{{11} \over {18}}1811​
  4. D
    1318{{13} \over {18}}1813​
View written solutionFree

Correct answer: D

  1. Identify the conic from the parametric form

Given x=2t,y=t23.x=2t,\qquad y=\frac{t^2}{3}.x=2t,y=3t2​.

Eliminate ttt: t=x2t=\frac{x}{2}t=2x​ so y=13(x2)2=x212.y=\frac{1}{3}\left(\frac{x}{2}\right)^2=\frac{x^2}{12}.y=31​(2x​)2=12x2​.

Thus the conic is the parabola x2=12y.x^2=12y.x2=12y.

Comparing with the standard form x2=4ay,x^2=4ay,x2=4ay, we get 4a=12  ⟹  a=3.4a=12 \implies a=3.4a=12⟹a=3.

Hence the focus is S=(0,3),S=(0,3),S=(0,3), and the axis is the yyy-axis.

  1. Coordinates of point AAA on the parabola

For parameter ttt, A=(2t,t23).A=(2t,\tfrac{t^2}{3}).A=(2t,3t2​).

  1. Find point BBB on the axis such that SA⊥BASA \perp BASA⊥BA

Since BBB lies on the axis of the parabola, let B=(0,b).B=(0,b).B=(0,b).

Condition SA⊥BASA \perp BASA⊥BA means (AS→)⋅(AB→)=0.(\overrightarrow{AS})\cdot(\overrightarrow{AB})=0.(AS)⋅(AB)=0.

Now, AS→=S−A=(0−2t, 3−t23)=(−2t, 3−t23),\overrightarrow{AS}=S-A=(0-2t,\ 3-\tfrac{t^2}{3})=(-2t,\ 3-\tfrac{t^2}{3}),AS=S−A=(0−2t, 3−3t2​)=(−2t, 3−3t2​), AB→=B−A=(0−2t, b−t23)=(−2t, b−t23).\overrightarrow{AB}=B-A=(0-2t,\ b-\tfrac{t^2}{3})=(-2t,\ b-\tfrac{t^2}{3}).AB=B−A=(0−2t, b−3t2​)=(−2t, b−3t2​).

So, (−2t)(−2t)+(3−t23)(b−t23)=0.(-2t)(-2t)+\left(3-\frac{t^2}{3}\right)\left(b-\frac{t^2}{3}\right)=0.(−2t)(−2t)+(3−3t2​)(b−3t2​)=0.

That is, 4t2+(3−t23)(b−t23)=0.4t^2+\left(3-\frac{t^2}{3}\right)\left(b-\frac{t^2}{3}\right)=0.4t2+(3−3t2​)(b−3t2​)=0.

Since 3−t23=9−t23,3-\frac{t^2}{3}=\frac{9-t^2}{3},3−3t2​=39−t2​, we get (9−t23)(b−t23)=−4t2.\left(\frac{9-t^2}{3}\right)\left(b-\frac{t^2}{3}\right)=-4t^2.(39−t2​)(b−3t2​)=−4t2.

Thus b−t23=−12t29−t2.b-\frac{t^2}{3}=-\frac{12t^2}{9-t^2}.b−3t2​=−9−t212t2​.

Hence b=t23−12t29−t2.b=\frac{t^2}{3}-\frac{12t^2}{9-t^2}.b=3t2​−9−t212t2​.

Take LCM:

=t^2\cdot \frac{9-t^2-36}{3(9-t^2)} =-\frac{t^2(t^2+27)}{3(9-t^2)}.$$ So $$B=\left(0,-\frac{t^2(t^2+27)}{3(9-t^2)}\right).$$ --- 4. **Find the centroid of $\triangle SAB$** Coordinates of the vertices are: $$S=(0,3),\quad A=(2t,\tfrac{t^2}{3}),\quad B=(0,b).$$ The centroid is $$\left(\frac{0+2t+0}{3},\frac{3+\frac{t^2}{3}+b}{3}\right).$$ Its ordinate is $$k=\frac{3+\frac{t^2}{3}+b}{3}.$$ Substitute $b=\frac{t^2}{3}-\frac{12t^2}{9-t^2}$: $$k=\frac{3+\frac{t^2}{3}+\frac{t^2}{3}-\frac{12t^2}{9-t^2}}{3} =\frac{3+\frac{2t^2}{3}-\frac{12t^2}{9-t^2}}{3}.$$ Now take the limit as $t\to 1$: $$k\to \frac{3+\frac{2}{3}-\frac{12}{8}}{3} =\frac{3+\frac{2}{3}-\frac{3}{2}}{3}.$$ Compute inside: $$3+\frac{2}{3}=\frac{11}{3},$$ so $$\frac{11}{3}-\frac{3}{2}=\frac{22-9}{6}=\frac{13}{6}.$$ Therefore, $$\lim_{t\to 1} k=\frac{1}{3}\cdot \frac{13}{6}=\frac{13}{18}.$$ --- 5. **Check options** The value is $$\frac{13}{18},$$ which matches **Option D**. --- **Final Answer:** $\boxed{\frac{13}{18}}$
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