The line y=mx+1 is a tangent to the curve y^{ | vedclass

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(A) The equation of the tangent to the given curve is $y=mx+1$.Substituting $y=mx+1$ into the equation of the parabola $y^{2}=4x$,we get:$(mx+1)^{2}=4

Vedclass · Accepted Answer

$1$

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(A) The equation of the tangent to the given curve is $y=mx+1$.Substituting $y=mx+1$ into the equation of the parabola $y^{2}=4x$,we get:$(mx+1)^{2}=4x$Expanding the equation:$m^{2}x^{2}+2mx+1=4x$$m^{2}x^{2}+x(2m-4)+1=0......(i)$Since the line is a tangent to the curve,it touches the curve at exactly one point. Therefore,the quadratic equation $(i)$ must have equal roots,which implies that its discriminant $D$ must be zero.$D = b^{2}-4ac = 0$$(2m-4)^{2}-4(m^{2})(1)=0$Expanding the discriminant:$4m^{2}-16m+16-4m^{2}=0$$-16m+16=0$$16m=16$$m=1$Thus,the required value of $m$ is $1$.The correct answer is $A$.

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

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