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Parabola question

2015 · Shift 0 · Q36
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Parabola question

2015 · Shift 0 · Q36

JEE MainMathematicsParabolaMCQ+4 / −1
Let OOO be the vertex and QQQ be any point on the parabola, x2=8y{{x^2} = 8y}x2=8y. If the point PPP divides the line segment OQOQOQ internally in the ratio 1:31:31:3, then locus of PPP is :
  1. A
    y2=2x{y^2} = 2xy2=2x
  2. B
    x2=2y{{x^2} = 2y}x2=2y
  3. C
    x2=y{{x^2} = y}x2=y
  4. D
    y2=x{y^2} = xy2=x
View written solutionFree

Correct answer: B

  1. Write the parabola in standard parametric form

    Given parabola: x2=8yx^2=8yx2=8y

    Compare with the standard form: x2=4ayx^2=4ayx2=4ay so, 4a=8⇒a=24a=8 \Rightarrow a=24a=8⇒a=2

    A general point QQQ on the parabola x2=4ayx^2=4ayx2=4ay is Q=(2at,at2)Q=(2at,at^2)Q=(2at,at2) Hence here, Q=(4t,2t2)Q=(4t,2t^2)Q=(4t,2t2)

  2. Coordinates of the vertex

    The vertex OOO of the parabola is O=(0,0)O=(0,0)O=(0,0)

  3. Use the section formula

    Point PPP divides the line segment OQOQOQ internally in the ratio 1:31:31:3.

    This means if O=(0,0)O=(0,0)O=(0,0) and Q=(4t,2t2)Q=(4t,2t^2)Q=(4t,2t2), then OP:PQ=1:3OP:PQ=1:3OP:PQ=1:3

    So PPP is one-fourth of the way from OOO to QQQ.

    Therefore, P=(1⋅4t+3⋅01+3,1⋅2t2+3⋅01+3)P=\left(\frac{1\cdot 4t+3\cdot 0}{1+3},\frac{1\cdot 2t^2+3\cdot 0}{1+3}\right)P=(1+31⋅4t+3⋅0​,1+31⋅2t2+3⋅0​) or more simply, P=(4t4,2t24)=(t,t22)P=\left(\frac{4t}{4},\frac{2t^2}{4}\right)=(t,\frac{t^2}{2})P=(44t​,42t2​)=(t,2t2​)

  4. Eliminate the parameter

    Let coordinates of PPP be (x,y)(x,y)(x,y). Then x=t,y=t22x=t, \qquad y=\frac{t^2}{2}x=t,y=2t2​

    From x=tx=tx=t, we get t=xt=xt=x

    Substitute into yyy: y=x22y=\frac{x^2}{2}y=2x2​

    Hence, x2=2yx^2=2yx2=2y

  5. Match with the options

    The locus is x2=2yx^2=2yx2=2y

    So the correct option is B.

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