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Parabola question

2005 · Shift 0 · Q108
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Parabola question

2005 · Shift 0 · Q108

JEE MainMathematicsParabolaMCQ+4 / −1
Let PPP be the point (1,0)(1, 0)(1,0) and QQQ a point on the parabola y2=8x{y^2} = 8xy2=8x. The locus of mid point of PQPQPQ is :
  1. A
    y2−4x+2=0{y^2} - 4x + 2 = 0y2−4x+2=0
  2. B
    y2+4x+2=0{y^2} + 4x + 2 = 0y2+4x+2=0
  3. C
    x2+4y+2=0{x^2} + 4y + 2 = 0x2+4y+2=0
  4. D
    x2−4y+2=0{x^2} - 4y + 2 = 0x2−4y+2=0
View written solutionFree

Correct answer: A

  1. Coordinates of the moving point on the parabola

Let the point QQQ on the parabola y2=8xy^2=8xy2=8x be Q=(X,Y).Q=(X,Y).Q=(X,Y). Then it satisfies Y2=8X.Y^2=8X.Y2=8X.

  1. Midpoint of PQPQPQ

Given P=(1,0)P=(1,0)P=(1,0).

Let the midpoint of PQPQPQ be M=(x,y)M=(x,y)M=(x,y). Using the midpoint formula, x=1+X2,y=0+Y2=Y2.x=\frac{1+X}{2},\qquad y=\frac{0+Y}{2}=\frac{Y}{2}.x=21+X​,y=20+Y​=2Y​. So, X=2x−1,Y=2y.X=2x-1,\qquad Y=2y.X=2x−1,Y=2y.

  1. Use the parabola condition

Since QQQ lies on y2=8xy^2=8xy2=8x, we substitute X=2x−1X=2x-1X=2x−1 and Y=2yY=2yY=2y: (2y)2=8(2x−1).(2y)^2=8(2x-1).(2y)2=8(2x−1). This gives 4y2=16x−8.4y^2=16x-8.4y2=16x−8. Divide by 444: y2=4x−2.y^2=4x-2.y2=4x−2. Rearrange: y2−4x+2=0.y^2-4x+2=0.y2−4x+2=0.

  1. Match with the options

The locus is y2−4x+2=0,y^2-4x+2=0,y2−4x+2=0, which is Option A.

  1. Comparison with stored correct answer

Stored correct answer: A

Our derived answer: A

They match.

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