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

Open in App

Solution

$y = x^{2} + 4 x + 1$ Points $(- 1, - 2)$
Step $1$ :- $y^{'} = 2 x + 4$
$\frac{d y}{d x} (x = - 1) = 2 \times (- 1) + 4$
$= 2$
$m = 2$
Step $2$ :- Equation of tangent:
$y - y^{'} = m (x - x_{1})$
$y - (- 2) = 2 (x - (- 1))$
$y + 2 = 2 (x + 1)$
$y + 2 = 2 x + 2$
$y + 2 - 2 x - 2 = 0$
$- 2 x + y = 0$
$2 x - y = 0$
Step $3$ :- $m_{1} m_{2} = 7$
$m_{1} m_{2} = - 7$
$2 \times m_{2} = - 7$
$m_{2} = - \frac{7}{2}$
Step $4$ :- $y - (- 2) = - \frac{1}{2} (x - (- 1))$
$y + 2 = - \frac{1}{2} (x + 1)$ $∴$ Equation of tangent is $2 x - y = 0$
$2 y + 4 = - x = 1$ $∴$ Equation of normal is $x + 2 y + 5 = 0$
$2 y + 4 + x + 1 = 0$
$x + 2 y + 5 = 0$ .

Suggest Corrections

Similar questions

y = 2 x^{2} - 3 x - 1

(1, - 2)

y = x^{2} + 4 x + 1

(- 1, - 2)

2 x^{2} + 3 y^{2} - 5 = 0

(1, 1)

x = a {cos}^{3} θ, y = a {sin}^{3} θ a t θ = \frac{π}{4}

y = x^{2} + 4 x + 1

(- 1, - 2)

2 x^{2} + 3 y^{2} - 5 = 0

(1, 1)

x = a {cos}^{3} θ, y = a {sin}^{3} θ

θ = \frac{π}{4}

y = x^{2}

(0, 0)

x = \frac{2 a t^{2}}{1 + t^{2}}, y = \frac{2 a t^{3}}{1 + t^{2}}

t = 1 / 2

Join BYJU'S Learning Program