Question

The line $y=mx+1$ is a tangent to the curve $y^2=4x$ if the value of $m$ is ___________

Answer 1

Find out the value of $m$.

Substitute the equation of the line $y=mx+1$into the equation of the given curve.

$$\begin{gathered} \therefore {(mx+1)}^2=4x \\ \Rightarrow m^2{x}^2+2mx+1=4x\left[\because {(a+b)}^2=a^2+b^2+2ab\right] \\ \Rightarrow m^2{x}^2+(2m-4)x+1=0 \end{gathered}$$

As we know,

Tangent touches a curve at one point, Thus, the discriminant of the above quadratic equation should be zero.

$$\begin{gathered} \therefore {(2m-4)}^2-4m^2=0\left[\because D=b^2-4ac\right] \\ \Rightarrow 4m^2+16-16m-4m^2=0 \\ \Rightarrow m=\frac{16}{16} \\ \Rightarrow m=1 \end{gathered}$$

Hence, the value of $m$ is equal to $1$.