The abscissa of the point on the curve y=alef | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the curve equation: $y = a\left(e^{\frac{x}{a}} + e^{-\frac{x}{a}}\right)$.To find the slope of the tangent,we differentiate $y$ with respec

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given the curve equation: $y = a\left(e^{\frac{x}{a}} + e^{-\frac{x}{a}}\right)$.To find the slope of the tangent,we differentiate $y$ with respect to $x$:$\frac{dy}{dx} = a \left( e^{\frac{x}{a}} \cdot \frac{1}{a} + e^{-\frac{x}{a}} \cdot (-\frac{1}{a}) \right) = e^{\frac{x}{a}} - e^{-\frac{x}{a}}$.Since the tangent is parallel to the $X$-axis,its slope must be zero:$\frac{dy}{dx} = 0 \implies e^{\frac{x}{a}} - e^{-\frac{x}{a}} = 0$.$e^{\frac{x}{a}} = e^{-\frac{x}{a}}$.Taking the natural logarithm on both sides:$\frac{x}{a} = -\frac{x}{a} \implies \frac{2x}{a} = 0 \implies x = 0$.Thus,the abscissa of the point is $0$.

The abscissa 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

Similar Questions

If the curve $y = ax^{2} + bx + c, x \in R,$ passes through the point $(1, 2)$ and the tangent line to this curve at the origin is $y = x$,then the possible values of $a, b, c$ are:

If the curves $2x = y^2$ and $2xy = K$ intersect perpendicularly,then the value of $K^2$ is

What is the length of the sub-tangent at any point on the curve $y = ax^3$?

If the length of the subtangent at any point to the curve $x y^{n}=a$ is proportional to the abscissa,then $n$ is

The length of the subnormal at any point on the curve $y = (\frac{x}{2024})^k$ is constant if the value of $k$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module