The x-coordinate of the point on the curve y | vedclass

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(D) Given the curve equation: $y = a\left( {{e^{\frac{x}{a}}} + {e^{ - \frac{x}{a}}}} \right)$.Differentiating with respect to $x$:$\frac{dy}{dx} = a

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(D) Given the curve equation: $y = a\left( {{e^{\frac{x}{a}}} + {e^{ - \frac{x}{a}}}} \right)$.Differentiating with respect to $x$:$\frac{dy}{dx} = a \left( \frac{1}{a} e^{\frac{x}{a}} - \frac{1}{a} e^{-\frac{x}{a}} \right) = e^{\frac{x}{a}} - e^{-\frac{x}{a}}$.Since the tangent is parallel to the $x$-axis,its slope is $0$:$\frac{dy}{dx} = 0$.Therefore,$e^{\frac{x}{a}} - e^{-\frac{x}{a}} = 0$.$e^{\frac{x}{a}} = e^{-\frac{x}{a}}$.Taking the natural logarithm on both sides:$\frac{x}{a} = -\frac{x}{a}$.$2x = 0$,which implies $x = 0$.

The $x$-coordinate of the point on the curve $y = a\left( {{e^{\frac{x}{a}}} + {e^{ - \frac{x}{a}}}} \right)$ where the tangent is parallel to the $x$-axis is ........

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