Find the equation

Question:

Find the equation of the tangent and the normal to the following curves at the indicated points:

$y=x^{2}+4 x+1$ at $x=3$

Solution:

finding slope of the tangent by differentiating the curve

$\frac{d y}{d x}=2 x+4$

$\mathrm{m}$ (tangent) at $(3,0)=10$

normal is perpendicular to tangent so, $m_{1} m_{2}=-1$

$\mathrm{m}$ (normal) at $(3,0)=-\frac{1}{10}$

equation of tangent is given by $\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}$ (tangent) $\left(\mathrm{x}-\mathrm{x}_{1}\right)$

$y$ at $x=3$

$y=3^{2}+4 \times 3+1$

$y=22$

$y-22=10(x-3)$

$y=10 x-8$

equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$

$y-22=-\frac{1}{10}(x-3)$

$x+10 y=223$

Leave a comment