The coordinate of the point on $y^{2}$ = 8x which is closest from $x^{2} + (y + 6)^{2}$ = 1 is

(2, 4)

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(18, 12)

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(18, –12)

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(2, –4)

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Solution

The correct option is D
(2, –4)

Point on Parabola = $(2 t^{2}, 4 t)$

Normal is y + xt = 4t + $2 t^{3}$

(0,-6) \Rightarrow 4t + $2 t^{3}$ = -6

$\Rightarrow$ t = -1

$∴$ Point = (2,-4)

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The coordinate of the point on y2 = 8x which is closest from x2+(y+6)2= 1 is

The correct option is D (2, –4) Point on Parabola = (2t2,4t) Normal is y + xt = 4t + 2t3 (0,-6) \Rightarrow 4t + 2t3= -6 ⇒ t = -1 ∴ Point = (2,-4)

The coordinate of the point on $y^{2}$ = 8x which is closest from $x^{2} + (y + 6)^{2}$ = 1 is

The correct option is D
(2, –4)

Point on Parabola = $(2 t^{2}, 4 t)$

Normal is y + xt = 4t + $2 t^{3}$

(0,-6) \Rightarrow 4t + $2 t^{3}$ = -6

$\Rightarrow$ t = -1

$∴$ Point = (2,-4)