Find the point p on the parabola y = x² closest to the point (3, 0).

Solution:

Given, the equation of parabola is y = x² 

We have to find the point p on the parabola closest to the point (3, 0).

Let (x,y) be the point closest to the point (3, 0).

Using distance formula,

D = \(\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}\)

D = \(\sqrt{xt-3)^{2}+(y-0)^{2}}\)

D = \(\sqrt{(x-3)^{2}+(y)^{2}}\)

D = \(\sqrt{(x^{2}+9-6x+y^{2})}\)

Now substituting the value of y in the above equation,

D = \(\sqrt{(x^{2}+9-6x+(x^{2})^{2})}\)

\(D =\sqrt{(x^{2}+9-6x+x^{4})}\)

Differentiating with respect to x,

dD/dx = 4x³ + 2x - 6

To find the value of x, let dD/dx = 0

 4x³ + 2x - 6 = 0

Dividing by 2 on both sides,

2x³ + x - 3 = 0

On solving,

2x³ - 2x² + 2x² - 2x + 3x - 3 = 0

2x²(x - 1) + 2x(x - 1) + 3(x - 1) = 0

(x - 1)(2x² + 2x + 3) = 0

x - 1 = 0

x = 1

2x² + 2x + 3 ≠ 0

Second derivative,

d²D/dx² = 12x² + 2

12x² + 2 = 0

At x = 1,

d²D/dx² > 0

So, at x = 1, D is minimum.

Put the value of x in the equation of parabola,

So, y = (1)²

y = 1

Therefore, the point p is (1,1)

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Find the point p on the parabola y = x² closest to the point (3, 0).

Summary:

The point p on the parabola y = x² closest to the point (3, 0) is (1,1).