What is the equation of the normal line of $f(x)=x/(2-x^2)$ at $x = 3$ ?

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Answer 1 · 2018-05-02T13:00:41

$77y=-343x+996$

First we find the y-coordinates;
Substitute x=3
$3/(2-(3)^2)=-3/7$

Then differentiate $x/(2-x^2)$
$dy/dx$ = $[x^2+2]/(x^2-2)^2$

Substitute 3 to find the gradient of the tangent
$dy/dx$ = $[3^2+2]/[3^2-2]^2$
= $11/49$

General equation of normal:
$(y-y_1)$ =-1/(gradient of tangent) $(x-x_1)$

Substituting the values
$(y-(-3/7))=-1/(11/49)(x-3)$
$y+3/7=-49/11x+147/11$
$y=-49/11x+996/77$
$77y=-343x+996$

What is the equation of the normal line of f(x)=x/(2-x^2)f(x)=x/(2-x^2) at x = 3x = 3?