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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)

best_answer

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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Plabita

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