Question

Find the equation of tangent and normal to the curve = $3x^3-4x+7$ at the point whose abscissa is 1.

Answer 1

The equation of curve is $y=3x^2-4x+7$ ________ (1)

when $x=1$ $y=3-4+7=6$

$\therefore$ point with abscissa 1 is (1,6)

Diff. (1) w.r.t x, $\frac{dy}{dx}=6x-4$

At $(1,6)=6-4=2$

$\therefore$ slope of tangent =2

$\therefore$ equation of tangent at (1,6) is

$y-6=2(x-1)$

$y-6=2\times -2$

$\Rightarrow y=2x+4$

Also slope of Normal = $-\frac{1}{2}$

equation of Normal = $(y-6)=-\frac{1}{2}(x-1)$

$\Rightarrow 2y-12=-x+1$

$\Rightarrow x+2y-13=0$