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Parabola question

2021 · 24 Feb · Shift 1 · Q34
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  5. /2021 · 24 Feb · Shift 1 · Q34

Parabola question

2021 · 24 Feb · Shift 1 · Q34

JEE MainMathematicsParabolaMCQ+4 / −1
The locus of the mid-point of the line segment joining the focus of the parabola y2 = 4ax to a moving point of the parabola, is another parabola whose directrix is :
  1. A
    x = 0
  2. B
    x = - a2{a \over 2}2a​
  3. C
    x = a
  4. D
    x = a2{a \over 2}2a​
View written solutionFree

Correct answer: A

  1. Given parabola and its focus

    The parabola is y2=4axy^2 = 4axy2=4ax Its focus is S(a,0).S(a,0).S(a,0).

  2. Take a general point on the parabola

    A standard parametric point on y2=4axy^2=4axy2=4ax is P(at2,2at).P(at^2, 2at).P(at2,2at).

  3. Find the midpoint of SPSPSP

    Let the midpoint be M(h,k)M(h,k)M(h,k). Then h=a+at22=a(1+t2)2,h = \frac{a + at^2}{2} = \frac{a(1+t^2)}{2},h=2a+at2​=2a(1+t2)​, k=0+2at2=at.k = \frac{0+2at}{2} = at.k=20+2at​=at.

  4. Eliminate the parameter ttt

    From k=at  ⟹  t=ka,k = at \implies t = \frac{k}{a},k=at⟹t=ak​, substitute into hhh:

    = \frac{a}{2} + \frac{k^2}{2a}.$$ Rearranging, $$k^2 = 2a\left(h-\frac{a}{2}\right).$$ Replacing $(h,k)$ by $(x,y)$, the locus is $$y^2 = 2a\left(x-\frac{a}{2}\right).$$

  5. Identify this parabola

    Compare with standard form y2=4A(x−c).y^2 = 4A(x-c).y2=4A(x−c). Here, 4A=2a  ⟹  A=a2,c=a2.4A = 2a \implies A = \frac{a}{2}, \qquad c = \frac{a}{2}.4A=2a⟹A=2a​,c=2a​.

    For a parabola y2=4A(x−c),y^2 = 4A(x-c),y2=4A(x−c), the directrix is x=c−A.x = c-A.x=c−A.

    Therefore, x=a2−a2=0.x = \frac{a}{2} - \frac{a}{2} = 0.x=2a​−2a​=0.

  6. Check options

    • A: x=0x=0x=0 ✅
    • B: x=−a2x=-\dfrac{a}{2}x=−2a​ ❌
    • C: x=ax=ax=a ❌
    • D: x=a2x=\dfrac{a}{2}x=2a​ ❌

Therefore, the directrix of the locus parabola is x=0.\boxed{x=0}.x=0​.

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