If y=mx+c is the normal at a point on the parabola y2=8x whose focal distance is 8 units, then \small \left | c \right | is equal to :

 

  • Option 1)

    2\sqrt{3}

  • Option 2)

    8\sqrt{3}

  • Option 3)

    10\sqrt{3}

  • Option 4)

    16\sqrt{3}

 

Answers (1)

As we learnt in

Parametric coordinates of parabola -

x= at^{2}

y= 2at

- wherein

For the parabola.

y^{2}=4ax

 

 and

Focal distance -

The distance of a point on the parabola from the focus.

- wherein

 

 Here y^2=4ax\: ; \:a=2

Also given that distance between P(at^2,2at) \:and \: S=(a,0) \: is\:8\:units

Thus (at^2-a)^2 +4a^2t^2=64

a(t^2+1) = 8

Thus  t^2+1 = 4\Rightarrow t=\sqrt{3}

C= 2at(2+t^2)

Put value a=2, t=\sqrt{3}

\left | C \right |= 10\sqrt{3}


Option 1)

2\sqrt{3}

Incorrect

Option 2)

8\sqrt{3}

Incorrect

Option 3)

10\sqrt{3}

Correct

Option 4)

16\sqrt{3}

Incorrect

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