Question

$y=mx+1$ be a tangent to $y^2=4x$ then find the value of $m$.

Answer 1

Now $y=mx+1$ be a tangent to the parabola $y^2=4x$.

Then $(mx+1{)}^2=4x$ or $m^2{x}^2+x(2m-4)x+1=0$......(1) will have two equal roots.

Now (1) will have two equal root if

$(2m-4{)}^2-4m^2=0$

or, $4m^2-16m+16-4m^2=0$

or, $-16m+16=0$

or, $m=1$.