Find the equations of tangent and normal to the curves at the indicated points on it.
$y = x^{2} + 4 x + 1$ at $(- 1, - 2)$

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Solution

$y = x^{2} + 4 x + 1$
Equation of tangent at $(- 1, - 2) => T : (\frac{- 2 + y}{2}) = - x + \frac{4 (x - 1)}{2} + 1$
$=> T : y = 2 x$
Now, Let normal be equal to $N : y = m x + c$
$=> 2 (m) = 1$ (as $T$ perpendicular to $N$ )
$=> m = \frac{- 1}{2}$
Now, $N$ : $y = \frac{- 1}{2} x + c$
as $N$ passes through $(- 1, - 2)$ ,
$=> - 2 = \frac{- 1}{2} (- 1) + c$
$=> c = - \frac{5}{2}$
$=> N : y = - \frac{1}{2} x - \frac{5}{2}$
$=> N : 2 y + x + 5 = 0$ .

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