The line y = x + 3 has slope 1, so we look for points on the curve where the tangent line, whose slope is dy/dx, is equal to 1.
y² = x
Take the derivative of both sides with respect to x, assuming y = y(x) :
2y dy/dx = 1
dy/dx = 1/(2y)
Solve for y when dy/dx = 1 :
1 = 1/(2y)
2y = 1
y = 1/2
When y = 1/2, we have x = y² = (1/2)² = 1/4. However, for the given line, when y = 1/2, we have x = y - 3 = 1/2 - 3 = -5/2.
This means the line y = x + 3 is not a tangent to the curve y² = x. In fact, the line never even touches y² = x :
x = y² ⇒ y = y² + 3 ⇒ y² - y + 3 = 0
has no real solution for y.
