Question

Choose the correct answer in the following questions :
The line y=mx + 1 is a tangent to the curve $y^2=4x,$ if the value of m is
(a) $1$ (b) $2$ (c) $3$ (d) $\frac{1}{2}$

Answer 1

The equation of the tangent to the curve is y = mx + 1.
Slope of the given line is m, Given curve is $y^2=4x$ ...(i)
On differentiating, we get
$2y\frac{dy}{dx}=4\Rightarrow \frac{dy}{dx}=\frac{2}{y}$
If line y = mx + 1 is tangent of the curve, then we must have $\frac{2}{y}=m$
$\Rightarrow y=\frac{2}{m}$
Substituting this in Eq. (i), we get $x=\frac{y^2}{4}=\frac{1}{4}(\frac{2}{m}{)}^2=\frac{1}{m^2}$
$\therefore$ The point at which the given line touches Eq. (i) is ($\frac{1}{m^2},\frac{2}{m}$).
This point must lie on the line $y=mx+1.$
$\therefore \frac{2}{m}=m\left(\frac{1}{m^2}\right)+1\Rightarrow \frac{1}{m}=1\Rightarrow m=1$
Hence, the correct option is (a).