Determine k so that the line y = kx+1 is tangent

Determine $k$ so that the line $y = kx+1$ is tangent to the curve $y^{2} = 4x$, and find the point of tangency.

Step 1: Eliminate $x$

For $k\neq0$ the line gives $x = \dfrac{y-1}{k}$; substituting into $y^{2} = 4x$: \[y^{2} = \frac{4\bigl(y-1\bigr)}{k} \;\Longrightarrow\; ky^{2}-4y+4 = 0. \tag{1}\]

Step 2: Tangency

The line touches the parabola exactly when (1) has a double root, i.e. when its discriminant vanishes: \[\Delta = 16-16k = 0 \;\Longrightarrow\; k = 1.\]

Step 3: The point

With $k = 1$, (1) becomes $\bigl(y-2\bigr)^{2} = 0$, so $y_{0} = 2$ and $x_{0} = \dfrac{y_{0}^{2}}{4} = 1$.

Step 4: Check by differentiation

\[ 2yy' = 4 \;\Longrightarrow\; y' = \frac{2}{y}, \qquad y'\bigl(1,2\bigr) = 1 = k \;\;\checkmark \]

(The excluded case $k = 0$ gives the horizontal line $y = 1$, which crosses the parabola once, at $x = \tfrac14$, without touching it.)

Step 5: Verification

\[\text{On the curve: } 2^{2} = 4 = 4\cdot1 \;\;\checkmark; \qquad \text{on the line: } 1\cdot1+1 = 2 \;\;\checkmark\]

\[\text{Slope: } y'\bigl(1,2\bigr) = \tfrac22 = 1 \;\;\checkmark\]

\[\text{For } k\gt 1 \text{ the line misses the parabola entirely; for } 0\lt k\lt 1 \text{ it cuts it twice} \;\;\checkmark\]