The line y=x+1 is a tangent to the curve y^{2 | vedclass

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Question

(A) The equation of the given curve is $y^{2}=4x$. Differentiating with respect to $x$,we have: $2y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \fra

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$(1,2)$

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(A) The equation of the given curve is $y^{2}=4x$. Differentiating with respect to $x$,we have: $2y \frac{dy}{dx} = 4 \Rightarrow \frac{dy}{dx} = \frac{2}{y}$. Therefore,the slope of the tangent to the given curve at any point $(x, y)$ is given by $\frac{dy}{dx} = \frac{2}{y}$. The given line is $y=x+1$,which is of the form $y=mx+c$. The slope of the line is $1$. The line $y=x+1$ is a tangent to the given curve if the slope of the line is equal to the slope of the tangent. Thus,we must have $\frac{2}{y} = 1 \Rightarrow y=2$. Substituting $y=2$ into the equation of the line $y=x+1$,we get $2 = x+1 \Rightarrow x=1$. Hence,the line $y=x+1$ is a tangent to the given curve at the point $(1, 2)$. The correct answer is $A$.

The line $y=x+1$ is a tangent to the curve $y^{2}=4x$ at the point

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