Question

An enemy fighter jet is flying along the curve given by y = x² + 2. A soldier is placed at (3, 2) wants to shoot down the jet when it is nearest to him. Then the nearest distance is

• A. 2 units
• B. \(\sqrt3\) units
• C. \(\sqrt5\) units
• D. \(\sqrt6\) units

Hint:

To find the nearest distance between a point and a curve, use the distance formula and minimize the squared distance function. Differentiating the squared distance function and finding the critical points gives the point where the distance is minimum. In this case, the minimum distance occurs when \( x = 1 \), and the distance is \( \sqrt{5} \) units.

Answer 1

The Correct Option is C

The correct answer is: (C) \(\sqrt{5}\) units.

An enemy fighter jet is flying along the curve given by:

\(y = x^2 + 2\)

A soldier is placed at the point \( (3, 2) \) and wants to shoot down the jet when it is nearest to him. We are tasked with finding the nearest distance from the soldier to the jet. 
Step 1: Find the distance between the soldier and the jet

The distance \( d \) between a point \( (x, y) \) and the point \( (3, 2) \) is given by the distance formula: \[ d = \sqrt{(x - 3)^2 + (y - 2)^2} \] Substituting \( y = x^2 + 2 \) into this formula: \[ d = \sqrt{(x - 3)^2 + (x^2 + 2 - 2)^2} \] Simplifying: \[ d = \sqrt{(x - 3)^2 + x^4} \] 
Step 2: Minimize the distance function

To find the point where the distance is minimum, we differentiate \( d^2 \) with respect to \( x \) and set it equal to 0. Minimizing \( d^2 \) is equivalent to minimizing \( d \), since \( d \) is always positive. First, compute \( d^2 \): \[ d^2 = (x - 3)^2 + x^4 = x^2 - 6x + 9 + x^4 \] Now, differentiate \( d^2 \) with respect to \( x \): \[ \frac{d}{dx} (d^2) = 2x - 6 + 4x^3 \] Set the derivative equal to 0 to find the critical points: \[ 2x - 6 + 4x^3 = 0 \] \[ 4x^3 + 2x - 6 = 0 \] Divide through by 2: \[ 2x^3 + x - 3 = 0 \] 
Step 3: Solve for \( x \)

Using trial and error or numerical methods, we find that \( x = 1 \) satisfies the equation. 
Step 4: Calculate the distance when \( x = 1 \)

Now, substitute \( x = 1 \) into the distance formula: \[ d = \sqrt{(1 - 3)^2 + (1^2 + 2 - 2)^2} = \sqrt{(-2)^2 + (1)^2} = \sqrt{4 + 1} = \sqrt{5} \] Therefore, the nearest distance is \( \sqrt{5} \) units. 
Thus, the correct answer is (C) \(\sqrt{5}\) units.