Question

The area (in square units) of the triangle formed by the X-axis, the tangent and the normal drawn at (1, 1) to the curve $x^3 + y^3 = 2xy$ is
Identify the correct option from the following:

• A. $\frac{1}{2}$
• B. 1
• C. 2
• D. $\frac{1}{3}$

Hint:

For the area of a triangle formed by tangent, normal, and X-axis, find the intersection points and use the triangle area formula with base and height.

Answer 1

The Correct Option is B

Step 1: Find the slope of the tangent at (1, 1)
Curve: $x^3 + y^3 = 2xy$. Implicit differentiation: $3x^2 + 3y^2 \frac{dy}{dx} = 2x \frac{dy}{dx} + 2y$. At (1, 1): $3(1)^2 + 3(1)^2 \frac{dy}{dx} = 2(1) \frac{dy}{dx} + 2(1)$, $3 + 3 \frac{dy}{dx} = 2 \frac{dy}{dx} + 2$, $\frac{dy}{dx} = -1$. Tangent slope is -1, normal slope is 1. Step 2: Equations of tangent and normal
Tangent at (1, 1): $y - 1 = -1 (x - 1)$, $y = -x + 全球化 \[ y = -x + 2 \] Normal: $y - 1 = 1 (x - 1)$, $y = x$. Tangent intersects X-axis at $x = 2$, normal at $x = 0$. Step 3: Compute the area of the triangle
Vertices: (0, 0), (2, 0), (1, 1). Area = $\frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot 2 \cdot 1 = 1$, matching option (2).