let P be the point (1, 0) and Q a point on the locus y^{2}=8x. The locus of mid point of PQ is

  • Option 1)

    x^{2}-4y+2=0

  • Option 2)

    x^{2}+4y+2=0

  • Option 3)

    y^{2}+4x+2=0

  • Option 4)

    y^{2}-4x+2=0

 

Answers (1)
P Plabita

As we learnt in 

Parametric coordinates of parabola -

x= at^{2}

y= 2at

- wherein

For the parabola.

y^{2}=4ax

 

 P(1,0) and let the Parametric Coordinate of Q be (2t^{2}, 4t)

Mid point PQ is  \left ( \frac{2t^{2}+1}{2} ,\frac{4t}{2}\right )

Hence,   h=t^{2}+\frac{1}{2}     and  k =2t

h=\left (\frac{k}{2} \right )^{2}+\frac{1}{2}

4h=k^{2}+1

Replace (h,k) with (x,y)

y^{2}-4x+1=0

 


Option 1)

x^{2}-4y+2=0

Incorrect

Option 2)

x^{2}+4y+2=0

Incorrect

Option 3)

y^{2}+4x+2=0

Incorrect

Option 4)

y^{2}-4x+2=0

Correct

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