Question

If the tangent drawn at the point \((\alpha, \beta)\) on the curve \[ x^{2/3} + y^{2/3} = 4 \] is parallel to the line \[ \sqrt{3}x + y = 1, \] then \( \alpha^2 + \beta^2 =\)

• A. 10
• B. 9
• C. 28
• D. 19

Hint:

To find the point where the tangent is parallel to a given line, match the derivative (slope) of the curve with the slope of the line.

Answer 1

The Correct Option is C

We are given:
\[ x^{2/3} + y^{2/3} = 4 \tag{1} \] and the tangent is parallel to the line:
\[ \sqrt{3}x + y = 1 \Rightarrow \text{Slope} = -\sqrt{3} \tag{2} \] Step 1: Differentiate the curve implicitly
Differentiate both sides of equation (1) with respect to \(x\):
\[ \frac{d}{dx}(x^{2/3}) + \frac{d}{dx}(y^{2/3}) = 0 \Rightarrow \frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx} = 0 \] Solving for \(\frac{dy}{dx}\):
\[ \frac{2}{3}y^{-1/3} \cdot \frac{dy}{dx} = -\frac{2}{3}x^{-1/3} \Rightarrow \frac{dy}{dx} = -\left(\frac{y}{x}\right)^{1/3} \tag{3} \] Step 2: Use the slope condition
Given that the slope of the tangent is \(-\sqrt{3}\), equate it with equation (3):
\[ -\left(\frac{y}{x}\right)^{1/3} = -\sqrt{3} \Rightarrow \left(\frac{y}{x}\right)^{1/3} = \sqrt{3} \Rightarrow \frac{y}{x} = (\sqrt{3})^3 = 3\sqrt{3} \Rightarrow y = 3\sqrt{3}x \tag{4} \] Step 3: Substitute into the original curve
Substitute equation (4) into (1):
\[ x^{2/3} + (3\sqrt{3}x)^{2/3} = 4 \Rightarrow x^{2/3} + (27x^2)^{1/3} = 4 \Rightarrow x^{2/3} + 3x^{2/3} = 4 \Rightarrow 4x^{2/3} = 4 \Rightarrow x^{2/3} = 1 \Rightarrow x = 1 \] Now substitute back to find \(y\):
\[ y = 3\sqrt{3} \cdot 1 = 3\sqrt{3} \] Step 4: Calculate \(\alpha^2 + \beta^2\)
\[ \alpha^2 + \beta^2 = 1^2 + (3\sqrt{3})^2 = 1 + 27 = \boxed{28} \]