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Find the point on the curve y2 = 4x which is nearest to the point (2, –8).

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Given Curve is y2 = 4x …… (1) 

Let us assume the point on the curve which is nearest to the point (2, - 8) be (x, y) 

The (x, y) satisfies the relation(1) 

Let us find the distance(S) between the points (x, y) and (2, - 8) 

We know that distance between two points (x1,y1) and (x2,y2) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

⇒ S = \(\sqrt{(x-2)^2+(y+8)^2}\)

⇒ S = \(\sqrt{x^2-4x+4+y^2+16y+64}\)

⇒ S = \(\sqrt{x^2-4x+y^2+16y+68}\)

Squaring on both sides we get, 

⇒ S2 = x2 + y2 - 4x + 16y + 68

From (1)

We know that distance is an positive number so, for a minimum distance S, S2 will also be minimum 

Let us S2 as the function of y. 

For maxima and minima,

We get minimum distance at y = 4 

Let find the value of x at these y values 

⇒ x = \(\frac{(4)^2}{4}\)

⇒ x = 4 

∴ The nearest point to the point (2, - 8) on the curve is (4,4).

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