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Parabola question

2011 · Shift 2 · Q38
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Parabola question

2011 · Shift 2 · Q38

JEE AdvancedMathematicsParabolaMultiple correct+4 / −1
Let L be a normal to the parabola y2 = 4x. If L passes through the point (9, 6), then L is given by
  1. A
    y −-− x + 3 = 0
  2. B
    y + 3x −-− 33 = 0
  3. C
    y + x −-− 15 = 0
  4. D
    7 −-− 2x + 12 = 0
View written solutionFree

Correct answer: A, B, D

  1. Parametric point on the parabola

For the parabola y2=4x,y^2=4x,y2=4x, a general point can be written as (t2,2t).(t^2,2t).(t2,2t).

  1. Equation of normal to y2=4xy^2=4xy2=4x

For y2=4axy^2=4axy2=4ax, the normal at parameter ttt is y=−tx+2at+t3.y=-tx+2at+t^3.y=−tx+2at+t3. Here 4a=4⇒a=14a=4\Rightarrow a=14a=4⇒a=1, so the normal is y=−tx+2t+t3.y=-tx+2t+t^3.y=−tx+2t+t3.

  1. Normal passes through (9,6)(9,6)(9,6)

Substitute (x,y)=(9,6)(x,y)=(9,6)(x,y)=(9,6): 6=−9t+2t+t36=-9t+2t+t^36=−9t+2t+t3 6=t3−7t6=t^3-7t6=t3−7t t3−7t−6=0.t^3-7t-6=0.t3−7t−6=0.

Now factorize: t3−7t−6=(t−3)(t+1)(t+2)=0.t^3-7t-6=(t-3)(t+1)(t+2)=0.t3−7t−6=(t−3)(t+1)(t+2)=0. So the possible values of ttt are t=3,−1,−2.t=3,-1,-2.t=3,−1,−2.

Thus there are three normals.

  1. Find each normal

For t=3t=3t=3

y=−3x+2(3)+33=−3x+6+27=−3x+33y=-3x+2(3)+3^3=-3x+6+27=-3x+33y=−3x+2(3)+33=−3x+6+27=−3x+33 y+3x−33=0.y+3x-33=0.y+3x−33=0. This is Option B.

For t=−1t=-1t=−1

y=−(−1)x+2(−1)+(−1)3=x−2−1=x−3y=-(-1)x+2(-1)+(-1)^3=x-2-1=x-3y=−(−1)x+2(−1)+(−1)3=x−2−1=x−3 y−x+3=0.y-x+3=0.y−x+3=0. This is Option A.

For t=−2t=-2t=−2

y=−(−2)x+2(−2)+(−2)3=2x−4−8=2x−12y=-(-2)x+2(-2)+(-2)^3=2x-4-8=2x-12y=−(−2)x+2(−2)+(−2)3=2x−4−8=2x−12 y−2x+12=0.y-2x+12=0.y−2x+12=0. This is not exactly Option D as printed, but Option D appears to be a typographical error and is intended to be y−2x+12=0.y-2x+12=0.y−2x+12=0.

  1. Check options
  • A: y−x+3=0y-x+3=0y−x+3=0 ✅
  • B: y+3x−33=0y+3x-33=0y+3x−33=0 ✅
  • C: y+x−15=0y+x-15=0y+x−15=0 ❌
  • D: printed as 7−2x+12=07-2x+12=07−2x+12=0, which simplifies to −2x+19=0-2x+19=0−2x+19=0, not a normal line. But the intended option is clearly y−2x+12=0y-2x+12=0y−2x+12=0 ✅
  1. Final answer

The normals are given by: A, B, D (with D interpreted as y−2x+12=0).\boxed{A,\ B,\ D\text{ (with D interpreted as }y-2x+12=0\text{)}}.A, B, D (with D interpreted as y−2x+12=0)​.

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