Final answer:
The point P with x-coordinate 1/2 lies on the curve y = 2x². To find the coordinates of point Q where the normal to the curve at P intersects the curve, we need to find the equation of the normal at P and then find the x-coordinates of Q by solving the resulting quadratic equation. Substituting the x-coordinates of Q into the equation of the curve will give us the corresponding y-coordinates of Q.
Step-by-step explanation:
The equation of the curve is given as y = 2x². We are given that the point P has an x-coordinate of 1/2. To find the y-coordinate of P, we substitute x = 1/2 into the equation: y = 2(1/2)² = 2(1/4) = 1/2. So, the coordinates of point P are (1/2, 1/2).
To find the normal to the curve at P, we need to find the slope of the tangent line at P. The slope of a curve at a point is the derivative of the curve at that point. Taking the derivative of y = 2x², we get dy/dx = 4x.
At point P, where x = 1/2, the slope is 4(1/2) = 2. The normal to the curve at P will have a slope that is the negative reciprocal of the tangent line's slope. So the slope of the normal at P is -1/2.
To find the equation of the normal at P, we use the point-slope form of a line: y - y₁ = m(x - x₁), where (x₁, y₁) are the coordinates of P and m is the slope of the normal. Substituting the values, we get: y - 1/2 = -1/2(x - 1/2). Simplifying, we have y = -x + 3/4.
Now, we need to find the coordinates of point Q, where the normal intersects the curve. We substitute the equation of the normal into the equation of the curve: 2x² = -x + 3/4. Rearranging and simplifying, we have the quadratic equation: 2x² + x - 3/4 = 0.
We can solve this quadratic equation using factoring, completing the square, or the quadratic formula. The solutions to the quadratic equation are the x-coordinates of points Q. Once we have the x-coordinates, we can substitute them into the equation of the curve to find the corresponding y-coordinates of Q.