Q. 31

# Find the point on

Given Curve is x2 = 8y …… (1)

Let us assume the point on the curve which is nearest to the point (2, 4) be (x, y)

The (x, y) satisfies the relation(1)

Let us find the distance(S) between the points (x, y) and (2, 4)

We know that distance between two points (x1,y1) and (x2,y2) is .

Squaring on both sides we get,

S2 = x2 + y2 - 4x - 8y + 20

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 x.

For maxima and minima,

x3 - 64 = 0

On solving we get

x = 4

Now differentiating again

At x = 4

We get minimum distance at x = 4

Let find the value of y at these x values

y = 2

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

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