Answer :


A point is present on a curve y2 = 4x

The point is near to the point (2,-8)

Let us consider,

The co-ordinates of the point be P(x,y)

As the point P is on the curve, then y2 = 4x

The distance between the points is given by,

D2 = (x-2)2 +(y+8)2

D2 = x2-4x+4 + y2 + 64 + 16y

Substituting x in the distance equation

---- (2)

For finding the maximum/ minimum of given function, we can find it by differentiating it with y and then equating it to zero. This is because if the function Z(x) has a maximum/minimum at a point c then Z’(c) = 0.

Differentiating the equation (2) with respect to y:

---- (2)

[Since ]

To find the critical point, we need to equate equation (2) to zero.

y3 + 64 = 0

(y + 4) (y2 – 4y + 16) = 0

(y+4) = 0 (or) y2 – 4y + 16 = 0

y = -4

(as the roots of the y2 – 4y + 16 are imaginary)

Now to check if this critical point will determine the distance is mimimum, we need to check with second differential which needs to be positive.

Consider differentiating the equation (2) with y:

----- (3)

[Since ]

Now let us find the value of

As , so the Distance D2 is minimum at y = -4

Now substituting y in x, we have

So, the point P on the curve y2 = 4x is (4,-4) which is at nearest from the (2,-8)

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