Final answer:
To find the intersection point of the curve with the surface, we substitute the parametric equations into the surface equation, solve the resulting quadratic equation using the quadratic formula, find the appropriate t-value, substitute back to get the point (a, b, c), and finally calculate a² + 2b.
Step-by-step explanation:
The point of intersection of the space curve r(t) = <√2t, t²+ 1, 1-4t> with the surface x²+2y-z = 0 gives us the coordinates (a, b, c). To find the point of intersection, we substitute the expressions for x, y, and z from the space curve into the equation of the surface:
Substituting √2t for x, t²+1 for y, and 1-4t for z we get:
(√2t)² + 2(t²+1) - (1-4t) = 0
2t + 2t² + 2 - 1 + 4t = 0
2t² + 6t + 1 = 0
This is a quadratic equation, which can be solved using the quadratic formula: -b ± √b² - 4ac / 2a. Here, a = 2, b = 6, c = 1, thus we find the value of t that satisfies the equation. After finding the t-value, we substitute back into r(t) to find the point (a, b, c). The question then asks us to compute a² + 2b.
For the quadratic equation 2t² + 6t + 1 = 0, we can apply the quadratic formula:
t = <-b ± √b² - 4ac / (2a)>
Upon solving, we find the appropriate root that gives us the point of intersection when plugged back into r(t). Finally, we use the coordinates of that point to calculate a² + 2b, choosing the right option from the given alternatives A to E.