Equation of tangent on the curve Y=e-|x| at the point where it cuts th

1.	Equation of tangent on the curve Y=e-\|x\| at the point where it cuts the line x=1
Answer» Question : Equation of TANGENT on the curve $y = e {}^{ - \|<klux>X</klux>\| }$ at the point where it cuts the line x=1 Theory ; Equation of tangent to a given curve : LET y= f(x) be a curve and let p $(x_{1},y_{1})$ on it .Then , $\dfrac{dy}{dx} = m = \bf slope \: of \: tagent \: at \: p$ ⇒The equation of tangent at p $(x_{1},y_{1})$ to the curve y=f(x) is $\bf (y - y_{1}) = \frac{dy}{dx} (x - x_{1})$ Solution : Given : $y = e {}^{ - \|x\| }$ To find : we have to find equation of tangent at the point where it cuts the line x=1 _______________________________ $y = e {}^{ - \|x\| }$ Here x= 1 i.e \|x\| = x [ x >0] $\implies y = e {}^{ - x}$ Now Differentiate with respect to x $\dfrac{dy}{dx} = e {}^{ - x} \times - 1$ $\bf \dfrac{dy}{dx} (at \: x = 1)= \bf \dfrac{ - 1}{e} = m$ Now put the value of x = 1 in the curve equation $y = e {}^{ - 1} = \frac{1}{e}$ ⇒Point $(1,\dfrac{1}{e})$ Equation of the tangent $(1,\dfrac{1}{e})$ is $(y - \frac{1}{e} ) = \frac{1}{e}(x - 1)$ $\implies \frac{ey - 1}{e} = \frac{ - 1}{e} (x - 1)$ $\implies \bf x + ey - 2 = 0$ _______________________________ THEREFORE, the Equation of tangent on the curve $y = e {}^{ - \|x\| }$ at the point where it cuts the line x=1 is : ${\purple{\boxed{\large{\bold{x+ey-2=0}}}}}$