Question 2

Parabola y² = 16x, so a = 4, focus (4, 0). Point (64, 32) = (at², 2at) with t = 4. Tangent slope at parameter t is 1/t = 1/4. Perpendicular tangent has slope −4, so t₁ = −1/4. Point: x₁ = a t₁² = 4·(1/16) = 1/4, y₁ = 2a t₁ = −2. Distance to focus = x₁ + a = 1/4 + 4 = 17/4.

**Calculating parameters for JSON output** I need to solve the equation y' = 8/y. For y = 32, it results in 1/4. The perpendicular gives y = -2 at x = 1/4. The focus is set at (4, 0), and the distance calculation leads to sqrt((15/4)^2 + 4), simplifying to 17/4 and pointing to option C. The user requests a final JSON output, so I’ll ensure that only the JSON is provided without additional explanations.

**Solving a parabola problem** I need to focus on solving this math problem about a parabola defined by the equation y² = 16x. The tangent point is at (64, 32), and there's another tangent that's perpendicular to it. I plug in the values, confirm the focus, and ensure the point lies on the parabola. After calculating the tangent's slope, I determine a new point. Ultimately, I find the distance to the focus is 17/4, which matches option C. **Calculating focal distance** I'm figuring out the focal distance from a point related to the parabola y² = 4ax. For this equation, the distance to the focus seems to be calculated as x + a, which equals a(1 + t²). So, plugging the numbers in, I find 4(1 + 1/16) gives me 17/4. It looks like option C is the right answer! Now I just need to make sure my final output is in JSON format without any extras.

For the parabola y²=16x, I find a=4, and the tangent slope at (64,32) is 1/4, so I need the perpendicular slope of -4. Setting 8/y1 = -4 gives y1=-2, then solving y1²=16x1 gives x1=1/4, so the point is (1/4, -2), and I'm noting the focus is at (4,0). Computing the distance between these points: sqrt((4-1/4)² + 4) = sqrt(225/16 + 64/16) = sqrt(289/16) = 17/4.