The curve defined implicitly by x^2+xy+y^2=3 passes through the point (1,1). (i) Find dy/dx at (1,1) using implicit differentiation. (ii) State whether the tangent at this point is horizontal, vertical, or neither.

The curve defined implicitly by $x^{2} + x y + y^{2} = 3$ passes through the point $(1, 1)$ . (i) Find $\frac{d y}{d x}$ at $(1, 1)$ using implicit differentiation. (ii) State whether the tangent at this point is horizontal, vertical, or neither.

hard 4 marks Class 12 · Mathematics · Continuity and Differentiability

Answer

(i)

- 1

\n(ii) Neither

The correct answer is (i) $- 1$ \n(ii) Neither — Hence (i) $- 1$ and (ii) neither horizontal nor vertical.

Differentiate both sides with respect to $x$ :\n $\frac{d}{d x} (x^{2}) = 2 x$ ,\n $\frac{d}{d x} (x y) = y + x \frac{d y}{d x}$ (product rule),\n $\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}$ .\nThus $2 x + y + x \frac{d y}{d x} + 2 y \frac{d y}{d x} = 0$ .
Collect terms containing $\frac{d y}{d x}$ : $(x + 2 y) \frac{d y}{d x} = - (2 x + y)$ .
Solve for $\frac{d y}{d x}$ : $\frac{d y}{d x} = \frac{- ( 2 x + y )}{x + 2 y}$ .
Substitute the point $(1, 1)$ : $\frac{d y}{d x} = \frac{- ( 2 \cdot 1 + 1 )}{1 + 2 \cdot 1} = \frac{- 3}{3} = - 1$ .
Since the slope $- 1$ is finite and non‑zero, the tangent line is neither horizontal nor vertical.
Hence (i) $- 1$ and (ii) neither horizontal nor vertical.

Common mistake: Missing the product rule for

x y

or plugging wrong values

Concept tested: Implicit differentiation and tangent nature