Find the point on the parabola y² = 2x that is closest to the point (1, 4)?

Solution:

Given, the equation of parabola is y² = 2x

Now differentiate with respect to x

2y dy/dx = 2

dy/dx = 1/y

We know that 

Slope of normal = -y

Now the equation of normal through the point (1, 4) is y - 4 = - y (x - 1)

At the intersection x = y² /2

y - 4 = - y (y² /2 - 1)

So we get

y - 4 = -y³ /2 + y

y³ = 8

y = 2

As x = y² /2 and y = 2 we get x = 2

Therefore, the point p is (2, 2)

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Find the point on the parabola y² = 2x that is closest to the point (1, 4)?

Summary:

The point p on the parabola y² = 2x closest to the point (1, 4) is (2, 2).