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(B) The given equation of the parabola is $y^2=4ax$. Let the equation of the line be $y=mx+c$. For the line to be a tangent to the parabola,the condit

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$x-my+am^2=0$

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(B) The given equation of the parabola is $y^2=4ax$. Let the equation of the line be $y=mx+c$. For the line to be a tangent to the parabola,the condition is $c = \frac{a}{m}$. Substituting this into the line equation,we get $y = mx + \frac{a}{m}$. Multiplying by $m$,we get $my = m^2x + a$. Replacing $m$ with $\frac{1}{m}$,we get $y = \frac{1}{m}x + am^2$,which simplifies to $my = x + am^2$. Rearranging the terms,we get $x - my + am^2 = 0$.

The line among the following which touches the parabola $y^2=4ax$ is

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