Question

The slope of tangent line to the curve 4x²+2xy+y²=12 at the point (1, 2) is

• A. 2
• B. 1
• C. -2
• D. -1
• E. 0

Answer 1

The Correct Option is D

We are given the equation of the curve \( 4x^2 + 2xy + y^2 = 12 \) and need to find the slope of the tangent line at the point \( (1, 2) \).

To find the slope of the tangent line, we need to compute \( \frac{dy}{dx} \) using implicit differentiation.

Differentiate both sides of the equation \( 4x^2 + 2xy + y^2 = 12 \) with respect to \( x \): 

\( \frac{d}{dx}(4x^2) + \frac{d}{dx}(2xy) + \frac{d}{dx}(y^2) = \frac{d}{dx}(12) \).

Now, apply the derivative rules:

• For \( 4x^2 \), \( \frac{d}{dx}(4x^2) = 8x \).
• For \( 2xy \), we apply the product rule: \( \frac{d}{dx}(2xy) = 2x \frac{dy}{dx} + 2y \).
• For \( y^2 \), apply the chain rule: \( \frac{d}{dx}(y^2) = 2y \frac{dy}{dx} \).
• The derivative of the constant \( 12 \) is 0.

Thus, the differentiated equation is:

\( 8x + 2x \frac{dy}{dx} + 2y \frac{dy}{dx} + 2y \frac{dy}{dx} = 0 \).

Simplify the equation:

\( 8x + \left( 2x + 2y \right) \frac{dy}{dx} = 0 \).

Now, solve for \( \frac{dy}{dx} \):

\( \frac{dy}{dx} = \frac{-8x}{2x + 2y} = \frac{-8x}{2(x + y)} \).

Substitute \( x = 1 \) and \( y = 2 \) into this expression to find the slope at the point \( (1, 2) \):

\( \frac{dy}{dx} = \frac{-8(1)}{2(1 + 2)} = \frac{-8}{6} = -\frac{4}{3} \).

The correct answer is \( -1 \).