Find an equation of the straight line normal to the curve y = 1/x at the point where x=a

Find an equation of the straight line normal to the curve y = 1/x at the point where x=a

y' = - 1/x^2

are the slope -1 ? by that i mean the tangent

Slope of the normal = -1/slope of the tangent

>> are the slope -1 ? <<
No. when x = a, the slope is -1 / a^2

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y = 1/x

y ' = -1 / x^2 <=== slope at x

-1 / a^2 <=== slope at a (tangent)
a^2 <=== slope of normal

y = 1/x
y = 1/a at x = a

y = mx + b

1/a = a^2 * a + b
b = 1/a - a^3

y = a^2 x + 1/a - a^3 <===== normal to the curve at x = a

y = 1/x
y(a) = 1/a

y` = -1/x²
y`(a) = -1/a²=m

y-y(a)=-(1/m)(x-a)
y-1/a=-(1/[-1/a²])(x-a)
y-1/a=a²(x-a)
y=a²x-a³+1/a