Find the points on the curve y = 8x2 closest to the point (0, 2).

Question

I have to get 2 points

Answer 1

Start by finding the distance between (0, 2) and (x, 8x²) on the curve y = 8x² using the distance formula.

s² = (x - 0)² + (8x² - 2)²

Simplify

s² = 64x⁴ - 31x² + 4

Differentiate with respect to x

2s ds/dx = 256x³ - 62x

ds/dx = (128x³ - 31x)/s

For extreme points, we can set ds/dx = 0 and get

(128x³ - 31x)/s = 0

128x(x² - 31/128) = 0

x = 0, ±(√62)/16

Plug these x in the formula for the distance:

x = 0:

s² = 64(0)⁴ - 31(0)² + 4 = 4

x = ±(√62)/16:

s² = 64(±(√62)/16)⁴ - 31(±(√62)/16)² + 4 = 0.246

The distance for x = 0 is greater than that for x = ±(√62)/16. Therefore x = 0 has a point of maximum distance from (0, 2) and x = ±(√62)/16 have points of minimum distance from (0, 2).

Now for x = 0, we have y = 8(0)² = 0 and for x = ±(√62)/16, we have y = 8(±(√62)/16)² = 31/16

The points ((√62)/16, 31/16) and (-(√62)/16, 31/16) on the curve y = 8x² are closest to point (0, 2).