Q. 244.0( 1 Vote )

# Find the point on

Answer :

Given,

• A point is present on a curve y^{2} = 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 y^{2} = 4x

• The distance between the points is given by,

D^{2} = (x-2)^{2} +(y+8)^{2}

D^{2} = x^{2}-4x+4 + y^{2} + 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.

y^{3} + 64 = 0

(y + 4) (y^{2} – 4y + 16) = 0

(y+4) = 0 (or) y^{2} – 4y + 16 = 0

y = -4

(as the roots of the y^{2} – 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 D^{2} is minimum at y = -4

Now substituting y in x, we have

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

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