equations are,
y = mx + 1 & y2 = 4x
By solving given equations we get
(mx + 1)2 = 4x
Expanding the above equation we get
![\[{{m}^{2}}{{x}^{2}}~+\text{ }2mx\text{ }+\text{ }1\text{ }=\text{ }4x\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f687dc2103769294a6d2498fd2dc4352_l3.png "Rendered by QuickLaTeX.com")
On rearranging we get
![\[\begin{array}{*{35}{l}} {{m}^{2}}{{x}^{2}}~+\text{ }2mx\text{ }-\text{ }4x\text{ }+\text{ }1\text{ }=\text{ }0 \\ m+{{x}^{2}}~+\text{ }x\text{ }\left( 2m\text{ }-\text{ }4 \right)\text{ }+\text{ }1\text{ }=\text{ }0 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-32cadb11993e59cd2d4407791423f241_l3.png "Rendered by QuickLaTeX.com")
As the line touches the parabola, above equation must have equal roots,
Discriminant (D) = 0
![\[\begin{array}{*{35}{l}} {{\left( 2m\text{ }-\text{ }4 \right)}^{2}}~\text{ }-4\text{ }\left( {{m}^{2}} \right)\text{ }\left( 1 \right)\text{ }=\text{ }0 \\ 4{{m}^{2}}~\text{ }-16m\text{ }+\text{ }16\text{ }\text{ }-4{{m}^{2}}~=\text{ }0 \\ -16\text{ }-m\text{ }+\text{ }16\text{ }=\text{ }0 \\ \text{ }m\text{ }+\text{ }1\text{ }=\text{ }0 \\ -m\text{ }=\text{ }1 \\ \end{array}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-292aa4015007aa6cae71681199f29e8e_l3.png "Rendered by QuickLaTeX.com")
Hence, the required value of m is 1.