The line $y=m x+1$ is a tangent to the curve


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) $\frac{1}{2}$


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.



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