# The line $y=x+1$ is a tangent to the curve $y^{2}=4 x$ at the point

Question:

The line $y=x+1$ is a tangent to the curve $y^{2}=4 x$ at the point

(A) (1, 2)

(B) (2, 1)

(C) (1, −2)

(D) (−1, 2)

Solution:

The equation of the given curve is $y^{2}=4 x$.

Differentiating with respect to x, we have:

$2 y \frac{d y}{d x}=4 \Rightarrow \frac{d y}{d x}=\frac{2}{y}$

Therefore, the slope of the tangent to the given curve at any point (xy) is given by,

$\frac{d y}{d x}=\frac{2}{y}$

The given line is $y=x+1$ (which is of the form $y=m x+c$ )

∴ Slope of the line = 1

The line y = x + 1 is a tangent to the given curve if the slope of the line is equal to the slope of the tangent. Also, the line must intersect the curve.

Thus, we must have:

$\frac{2}{y}=1$

$\Rightarrow y=2$

Now, $y=x+1 \Rightarrow x=y-1 \Rightarrow x=2-1=1$

Hence, the line y = x + 1 is a tangent to the given curve at the point (1, 2).