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The line y=mx+1 is a tangent to the curv...

The line y=mx+1 is a tangent to the curve `y^2=4x` if the value of m is(A) 1 (B) 2(C) 3(D) 1/2.

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To determine the value of \( m \) for which the line \( y = mx + 1 \) is a tangent to the curve \( y^2 = 4x \), we will follow these steps: ### Step 1: Substitute the line equation into the curve equation We start by substituting \( y = mx + 1 \) into the curve equation \( y^2 = 4x \). \[ (mx + 1)^2 = 4x \] ...
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The line y=m x+1 is a tangent to the curve y^2=4x , if the value of m is (a) 1 (b) 2 (c) 3 (d) 1/2

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Knowledge Check

  • The line y=x+1 is a tangent to the curve y^(2)=4x at the point:

    A
    `(1, 2)`
    B
    `(2, 1)`
    C
    `(1, -2)`
    D
    `(-1, 2)`

  • The line y=x+1 is a tangent to the curve y^(2)=4x at the point :

    A
    `(1, 2)`
    B
    `(2, 1)`
    C
    `(1, -2)`
    D
    `(-1, 2)`

  • The line y=x+1 is tangent to the curve y^(2)=4x at the point :

    A
    `(1, 2)`
    B
    `(2, 1)`
    C
    `(1, -2)`
    D
    `(-1, 2)`

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