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).