The length intercepted by the curve y^2 = 4x | vedclass

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(C) The equation of the line is $y = mx + c$. Given $dy/dx = m = 1$.Since the line passes through $(0, 1)$,we have $1 = 1(0) + c$,which gives $c = 1$.

Vedclass · Accepted Answer

$0$

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(C) The equation of the line is $y = mx + c$. Given $dy/dx = m = 1$.Since the line passes through $(0, 1)$,we have $1 = 1(0) + c$,which gives $c = 1$.Thus,the equation of the line is $y = x + 1$.To find the intersection with the parabola $y^2 = 4x$,substitute $y = x + 1$ into the parabola equation:$(x + 1)^2 = 4x$$x^2 + 2x + 1 = 4x$$x^2 - 2x + 1 = 0$$(x - 1)^2 = 0$This gives $x = 1$. Substituting $x = 1$ into $y = x + 1$,we get $y = 2$.Since the line intersects the parabola at only one point $(1, 2)$,the line is tangent to the parabola.Therefore,the length intercepted by the curve on the line is $0$.

The length intercepted by the curve $y^2 = 4x$ on the line satisfying $dy/dx = 1$ and passing through the point $(0, 1)$ is given by

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