Question

Equation of tangent on the curve Y=e-|x| at the point where it cuts the line x=1

Answer 1

Question :

Equation of tangent on the curve $y = e {}^{ - |x| }$ 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

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$y = e {}^{ - |x| }$

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

$(y - \frac{1}{e} ) = \frac{1}{e}(x - 1)$

$\implies \frac{ey - 1}{e} = \frac{ - 1}{e} (x - 1)$

$\implies \bf x + ey - 2 = 0$

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Therefore, the Equation of tangent on the curve $y = e {}^{ - |x| }$ at the point where it cuts the line x=1 is :

${\purple{\boxed{\large{\bold{x+ey-2=0}}}}}$