Question

Find the equation of the tangent and the normal to the following curve at the indicated point.
$y=2x^2-3x-1$ at $(1,-2)$.

Answer 1

$y=2x^2-3x-1$ [ Given ]

Differentiate both sides w.r.t $x,$

$\Rightarrow$ $(x_1,y_1)=(1,-2)$

Slop of tangent, $m={\left(\frac{dy}{dx}\right)}_{(1,-2)}=4-3=1$

Equation of tangent is,

$y-y-1=m(x-x_1)$

$\Rightarrow$ $y+2=1(x-1)$

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

$\Rightarrow$ $x-y-3=0$

Equation of normal is,

$y-y_1=\frac{-1}{m}(x-x_1)$

$\Rightarrow$ $y+2=-1(x-1)$

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

$\Rightarrow$ $x+y+1=0$

$\therefore$ Equation of tangent is $x-y-3=0$

$\therefore$ Equation of normal is $x+y+1=0$