Question

If $P$ is a point on the parabola $y=x^{2}+4$ which is closest to the straight line $y =4 x -1$, then the co-ordinates of $P$ are

• A. (3,13)
• B. (1,5)
• C. (-2,8)
• D. (2,8)

Answer 1

The Correct Option is D

To find the point on the parabola \(y = x^{2} + 4\) that is closest to the line \(y = 4x - 1\), we need to use the concept of the distance between a point and a line in the coordinate plane. 

1. Identify the general point \(P\) on the parabola. The parabola is given by the equation \(y = x^2 + 4\). Therefore, any point \(P\) on the parabola can be represented as \((x, x^2 + 4)\).
2. Find the distance from point \(P(x, x^2 + 4)\) to the line \(y = 4x - 1\). The formula for the distance \(d\) from a point \((x_1, y_1)\) to a line of the form \(ax + by + c = 0\) is:

\(d = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}\)

1. For the line \(y = 4x - 1\), rearrange it into standard form \(4x - y - 1 = 0\). Here, \(a = 4\), \(b = -1\), and \(c = -1\).
2. Substitute the point \(P(x, x^2 + 4)\) into the distance formula:

\(d = \frac{|4x - (x^2 + 4) - 1|}{\sqrt{4^2 + (-1)^2}} = \frac{|4x - x^2 - 5|}{\sqrt{17}}\)

1. To minimize the distance, minimize the numerator \(|4x - x^2 - 5|\). This is equivalent to minimizing the expression inside the absolute value, \(4x - x^2 - 5\).
2. Set \(f(x) = 4x - x^2 - 5\). To find the critical points, compute the derivative \(f'(x)\) and set it to zero:

\(f'(x) = 4 - 2x\)

1. Set \(f'(x) = 0\):

\(4 - 2x = 0 \Rightarrow x = 2\)

1. Substitute \(x = 2\) back into the parabola's equation to find the \(y\)-coordinate of the point:

\(y = 2^2 + 4 = 4 + 4 = 8\)

1. Thus, the point \(P(2, 8)\) is the point on the parabola closest to the line.
2. Verify by considering the options given: Among the options \((3,13)\), \((1,5)\), \((-2,8)\), and \((2,8)\), only \((2,8)\) is found to be the point closest to the line based on the calculations.

Therefore, the correct answer is (2, 8).