Final answer:
The point on the curve y = √(-1+2x) where the tangent is perpendicular to the line 12x+4y= 1 is found by equating the derivative of the curve to the negative reciprocal of the slope of the given line, leading to the point [5/9, 1/3].
Step-by-step explanation:
To find the point on the curve y = √(-1+2x) where the tangent is perpendicular to the line 12x+4y= 1, we first need to find the slope of the given line. This can be done by rewriting the equation in slope-intercept form (y=mx+b) to identify the slope (m).
The equation of the given line is 12x+4y= 1, which can be rewritten as y = -3x + 0.25, so the slope, m, is -3. A line perpendicular to this would have a slope that is the negative reciprocal, which is 1/3.
Next, we need to find the derivative of the curve y = √(-1+2x), which represents the slope of the tangent at any point.
Using the chain rule, the derivative of y with respect to x (dy/dx) is 1/(2√(-1+2x)) × 2, simplifying to 1/√(-1+2x).
This slope must equal 1/3 for the point we are seeking. Therefore, we solve:
1/√(-1+2x) = 1/3
Squaring both sides, we get (-1+2x) = 1/9, and solving for x gives x = 5/9. Substituting back into the original equation for y, we get y = √(-1+2×(5/9)) = √(1/9) = 1/3.
Therefore, the point on the curve where the tangent is perpendicular to 12x+4y= 1 is [5/9, 1/3].