Question: "What is the equation of the line that..."

Breanna Fisher

Answered question

2022-04-09

What is the equation of the line that is normal to

f (x) = - e^{x}

at x=-2?

maggionmoo

Beginner2022-04-10Added 16 answers

Let us recall that, f'(X) is the slope of tgt. to the curve C:y=f(x) at any pt.
(x,y).
For,

f (x) = - e^{x}, f^{'} (x) = - e^{x} \Rightarrow f^{'} (- 2) = - e^{- 2} = - \frac{1}{e^{2}}

∴

the slope of tgt. to C at x=-2 is

- \frac{1}{e^{2}}

, &, since, normal is

⊥

to tgt,. the slope normal atx=-2 will be

- \frac{1}{- \frac{1}{e^{2}}} = e^{2}

Also,

x = - 2 \Rightarrow f (- 2) = - e^{- 2} = - \frac{1}{e^{2}}

, so, the pt. of contact is

(- 2, - \frac{1}{e^{2}})

All together, the normal passes thro. pt.

(- 2, - \frac{1}{e^{2}})

and has slope

= e^{2}

. This gives us the eqn. of normal as :

y + \frac{1}{e^{2}} = e^{2} (x + 2)

,
or,

y e^{2} + 1 = x e^{4} + 2 e^{4}

, i.e. ,

x e^{4} - y e^{2} + 2 e^{4} - 1 = 0

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