Find the equation of tangent and normal to the curve $x^{2} + 3 x y + y^{2} = 5$ at point $(1, 1)$ on it.

Open in App

Solution

$x^{2} + 3 x y + y^{2} = 5 a t (1, 1)$
Taking $\frac{d y}{d x} 2 x + 3 y + 3 x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0 \frac{d y}{d x} (3 x + 2 y) = - (2 x + 3 y) \frac{d y}{d x} = - \frac{(2 x + 3 y)}{(3 x + 2 y)} \frac{d y}{d x} (1, 1) = - \frac{(5)}{5} = - 1$
Equation of tangent $(x - 1) = - 1 (y - 1) x + y - 2 = 0$
Equation of normal
$(x - 1) = 1 (y - 1) x - 1 = y - 1 x - y = 0$

Suggest Corrections

Similar questions

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

(\sqrt{2} a, b) .

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

(- 1, - 2)

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

(1, 1)

x^{2} - 2 x y + 4 y = 0

\frac{- 3}{2}

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

(\sqrt{2} a, b) .

Join BYJU'S Learning Program