Question:
Find the equation of the tangent and the normal to the following curves at the indicated points:
$y=2 x^{2}-3 x-1$ at $(1,-2)$
Solution:
finding the slope of the tangent by differentiating the curve
$\frac{d y}{d x}=4 x-3$
$\mathrm{m}($ tangent $)$ at $(1,-2)=1$
normal is perpendicular to tangent so, $m_{1} m_{2}=-1$
$\mathrm{m}$ (normal) at $(1,-2)=-1$
equation of tangent is given by $y-y_{1}=m(\operatorname{tangent})\left(x-x_{1}\right)$
$y+2=1(x-1)$
$y=x-3$
equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$
$y+2=-1(x-1)$
$y+x+1=0$
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