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Parabola question

2021 · 24 Feb · Shift 2 · Q30
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  5. /2021 · 24 Feb · Shift 2 · Q30

Parabola question

2021 · 24 Feb · Shift 2 · Q30

JEE MainMathematicsParabolaMCQ+4 / −1
If P is a point on the parabola y = x2 + 4 which is closest to the straight line y = 4x −-− 1, then the co-ordinates of P are :
  1. A
    (−-− 2, 8)
  2. B
    (2, 8)
  3. C
    (1, 5)
  4. D
    (3, 13)
View written solutionFree

Correct answer: B

  1. Point on the parabola

A general point on the parabola y=x2+4y=x^2+4y=x2+4 is P(x,x^2+4).

  1. Distance from a point to the line

The given line is y=4x−1 ⇒ 4x−y−1=0.y=4x-1 \,\Rightarrow\, 4x-y-1=0.y=4x−1⇒4x−y−1=0.

Distance of point P(x,x2+4)P(x,x^2+4)P(x,x2+4) from the line 4x−y−1=04x-y-1=04x−y−1=0 is d=∣4x−(x2+4)−1∣42+(−1)2=∣4x−x2−5∣17.d=\frac{|4x-(x^2+4)-1|}{\sqrt{4^2+(-1)^2}}=\frac{|4x-x^2-5|}{\sqrt{17}}.d=42+(−1)2​∣4x−(x2+4)−1∣​=17​∣4x−x2−5∣​.

Since 17\sqrt{17}17​ is constant, we minimize ∣4x−x2−5∣.|4x-x^2-5|.∣4x−x2−5∣.

  1. Simplify the expression

4x−x2−5=−(x2−4x+5)=−((x−2)2+1).4x-x^2-5=-(x^2-4x+5)= -\big((x-2)^2+1\big).4x−x2−5=−(x2−4x+5)=−((x−2)2+1).

Hence, ∣4x−x2−5∣=(x−2)2+1.|4x-x^2-5|=(x-2)^2+1.∣4x−x2−5∣=(x−2)2+1.

So we need to minimize (x−2)2+1.(x-2)^2+1.(x−2)2+1.

  1. Find minimum

The minimum of (x−2)2+1(x-2)^2+1(x−2)2+1 occurs when x=2.x=2.x=2.

Then y=x2+4=22+4=8.y=x^2+4=2^2+4=8.y=x2+4=22+4=8.

So the closest point is P=(2,8).P=(2,8).P=(2,8).

  1. Check options
  • A: (−2,8)(-2,8)(−2,8)  not on parabola since (−2)2+4=8(-2)^2+4=8(−2)2+4=8, actually on parabola, but not closest.
  • B: (2,8)(2,8)(2,8)  correct.
  • C: (1,5)(1,5)(1,5)  on parabola, but not closest.
  • D: (3,13)(3,13)(3,13)  on parabola, but not closest.

Therefore, the correct option is B.

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