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

1

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2

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3

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\frac{1}{2}

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answer required

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Solution

The correct option is A $1$
Lets put $y = m x + 1$ in $y^{2} = 4 x$ we get
${(m x + 1)}^{2} = 4 x$
$\Rightarrow m^{2} x^{2} + 1 + 2 m x - 4 x = 0$
Tangent touches a curve at one point so descriminant of the equation should be zero.
$\Rightarrow {(2 m - 4)}^{2} - 4 \times 1 \times m^{2} = 0$
$\Rightarrow 4 m^{2} + 16 - 16 m - 4 m^{2} = 0$
$\Rightarrow m = 1$

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The line y=mx+1 is a tangent to the curve y2=4x, if the value of m is

The correct option is A 1Lets put y=mx+1 in y2=4x we get(mx+1)2=4x⇒m2x2+1+2mx−4x=0Tangent touches a curve at one point so descriminant of the equation should be zero. ⇒(2m−4)2−4×1×m2=0⇒4m2+16−16m−4m2=0⇒m=1

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