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 solve this problem, we need to find a point on the parabola \( y = x^2 + 4 \) that is closest to the line \( y = 4x - 1 \). This can be achieved by minimizing the distance between a point \((x, y)\) on the parabola and the line. 

The distance \( D \) from a point \((x_1, y_1)\) to a line \( Ax + By + C = 0 \) is given by:

\(D = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}\)

The line equation can be rewritten in standard form: \( 4x - y - 1 = 0 \).

1. For a point \( P(x, x^2 + 4) \) on the parabola, we use the line equation to express the distance:
2. To minimize the distance, we need to minimize the expression \(|x^2 - 4x + 5|\).
3. The expression \(x^2 - 4x + 5\) is a quadratic function, and we can find its minimum or maximum value by completing the square or using the vertex formula. The vertex of a parabola \(ax^2 + bx + c\) is at:
4. Substituting \(x = 2\) in the quadratic expression:
5. Thus, the minimum value \(1\) occurs when \( x = 2 \).
6. Calculate the corresponding \( y \)-coordinate of the point on the parabola:
7. Therefore, the coordinates of the point \( P \) on the parabola that is closest to the line are \((2, 8)\).

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