Question

What is the equation of the normal line of `f(x)= (1+2x)^2` at `x=1`?

Answer 1

12y + x - 109 = 0

Explanation:

To obtain equation of normal , y-b = m(x-a) . Require to find it's gradient m and a point on it (a , b )

differentiating f(x) will give gradient of tangent and evaluating f(x) for x = 1 will give a point on it.

differentiate using the `color(blue)(" chain rule ")`

f'(x) = 2(1 + 2x)`d/dx (1+2x) = 2(1+2x) .2 = 4(1+2x)`

and f'(1) 4(1+2) = 12 = m of tangent

now tangent and normal are perpendicular and so the product of their gradients will equal -1

If `m_1 color(black)(" is gradient of normal")`

then `m.m_1 = -1`

`rArr 12 xx m_1 = -1 → m_1 = -1/12`

f(1) = `(1+2)^2 = 9 rArrcolor(black)(" (a , b ) = ( 1 , 9 ) ")`

equation of normal: y- 9 =`-1/12 (x - 1 )`
multiplying by 12 to eliminate fraction ) gives

12y - 108 = - x + 1

`rArr 12y + x -109 = 0`