Find the equations of tangent and normal to the curves at the indicated points on it.
(i) $y = x^{2} + 4 x + 1$ at $(- 1, - 2)$
(ii) $2 x^{2} + 3 y^{2} - 5 = 0$ at $(1, 1)$
(iii) $x = a {cos}^{3} θ, y = a {sin}^{3} θ$ at $θ = \frac{π}{4}$

Open in App

Solution

i) $\frac{d y}{d x} = 2 x + 4$ at $(- 1, - 2) ⟹ m = 2$
Equation of tangent $y + 2 = 2 (x + 1) ⟹ 2 x - y = 0$
Equation of normal $y + 2 = \frac{- 1}{2} (x + 1) ⟹ x + 2 y + 5 = 0$
ii) $\frac{d y}{d x} = \frac{5 - 4 x}{6 y}$ at $(1, 1) ⟹ m = \frac{- 1}{6}$
Equation of tangent $(y - 1) = \frac{- 1}{6} (x - 1) ⟹ x + 6 y - 7 = 0$
Equation of normal $(y - 1) = 6 (x - 1) ⟹ 6 x - y - 5 = 0$
iii) $\frac{d y}{d x} = - tan θ$ at $θ = \frac{π}{4} ⟹ m = - 1$
Equation of tangent $(y - \frac{a}{2 \sqrt{2}}) = - 1 (x - \frac{a}{2 \sqrt{2}}) ⟹ \sqrt{2} (x + y) = a$
Equation of normal $y - \frac{a}{2 \sqrt{2}} = x - \frac{a}{2 \sqrt{2}} ⟹ x - y = 0$

Suggest Corrections

Similar questions

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}

x = a {sin}^{3} θ

y = a {cos}^{3} θ

θ = \frac{π}{4} .

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

(- 1, - 2)

y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5

(0, 5)

y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5

(1, 3)

y = x^{3}

(1, 1)

y = x^{2}

(0, 0)

x = cos t, y = sin t

t = \frac{π}{4}

x = a {cos}^{3} θ, y = a {sin}^{3} θ

θ = \frac{π}{4}

Join BYJU'S Learning Program