Question:
The line $y=m x+1$ is a tangent to the curve $y^{2}=4 x$ if the value of $m$ is
(A) 1
(B) 2
(C) 3
(D) $\frac{1}{2}$
Solution:
The equation of the tangent to the given curve is $y=m x+1$.
Now, substituting $y=m x+1$ in $y^{2}=4 x$, we get:
$\Rightarrow(m x+1)^{2}=4 x$
$\Rightarrow m^{2} x^{2}+1+2 m x-4 x=0$
$\Rightarrow m^{2} x^{2}+x(2 m-4)+1=0$ ...(1)
Since a tangent touches the curve at one point, the roots of equation (i) must be equal.
Therefore, we have:
Discriminant $=0$
$(2 m-4)^{2}-4\left(m^{2}\right)(1)=0$
$\Rightarrow 4 m^{2}+16-16 m-4 m^{2}=0$
$\Rightarrow 16-16 m=0$
$\Rightarrow m=1$
Hence, the required value of m is 1.
The correct answer is A.