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(B) For the curve $y=e^{x}$,the slope of the tangent at $(c, e^{c})$ is given by $\frac{dy}{dx} = e^{x} \implies m_{t} = e^{c}$.The equation of the ta

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$4$

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(B) For the curve $y=e^{x}$,the slope of the tangent at $(c, e^{c})$ is given by $\frac{dy}{dx} = e^{x} \implies m_{t} = e^{c}$.The equation of the tangent at $(c, e^{c})$ is $y - e^{c} = e^{c}(x - c)$.To find the intersection with the $x$-axis,set $y=0$: $-e^{c} = e^{c}(x - c) \implies -1 = x - c \implies x = c - 1$.For the parabola $y^{2}=4x$,differentiating gives $2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{2}{y}$.At the point $(1,2)$,the slope of the tangent is $m = \frac{2}{2} = 1$.The slope of the normal is $m_{n} = -\frac{1}{m} = -1$.The equation of the normal at $(1,2)$ is $y - 2 = -1(x - 1) \implies y - 2 = -x + 1 \implies x + y = 3$.To find the intersection with the $x$-axis,set $y=0$: $x = 3$.Since the points of intersection on the $x$-axis are the same,we equate the $x$-coordinates: $c - 1 = 3 \implies c = 4$.

If the tangent of the curve $y=e^{x}$ at a point $(c, e^{c})$ and the normal to the parabola $y^{2}=4x$ at the point $(1,2)$ intersect at the same point on the $x$-axis,then the value of $c$ is

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