Question

If for $ \theta \in \left[ -\frac{\pi}{3}, 0 \right] $, the points $ (x, y) = \left( 3 \tan\left( \theta + \frac{\pi}{3} \right), 2 \tan\left( \theta + \frac{\pi}{6} \right) \right) $ lie on $ xy + \alpha x + \beta y + \gamma = 0 $, then $ \alpha^2 + \beta^2 + \gamma^2 $ is equal to:

• A. 80
• B. 72
• C. 96
• D. 75

Hint:

When solving problems involving trigonometric identities and equations, carefully match the coefficients of terms when simplifying the expression.

Answer 1

The Correct Option is D

Step 1: Understand the Parametric Equations

The points are given by:
\[ x = 3 \tan\left( \theta + \frac{\pi}{3} \right) \]
\[ y = 2 \tan\left( \theta + \frac{\pi}{6} \right) \]

Step 2: Use Trigonometric Identities

Let:
\[ A = \theta + \frac{\pi}{3} \]
\[ B = \theta + \frac{\pi}{6} \]

Thus, \( A - B = \frac{\pi}{6} \). Using the tangent of a difference formula:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]

Given \( \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}} \), we have:
\[ \frac{1}{\sqrt{3}} = \frac{\frac{x}{3} - \frac{y}{2}}{1 + \frac{x}{3} \cdot \frac{y}{2}} \]

Step 3: Simplify the Equation

Simplify numerator and denominator:
\[ \frac{1}{\sqrt{3}} = \frac{\frac{2x - 3y}{6}}{\frac{6 + xy}{6}} = \frac{2x - 3y}{6 + xy} \]

Multiply both sides by \(6 + xy\):
\[ \frac{6 + xy}{\sqrt{3}} = 2x - 3y \]

Multiply by \( \sqrt{3} \):
\[ 6 + xy = 2\sqrt{3}x - 3\sqrt{3}y \]

Rearrange terms:
\[ xy - 2\sqrt{3}x + 3\sqrt{3}y + 6 = 0 \]

Step 4: Compare with Given Curve

The given curve is:
\[ xy + \alpha x + \beta y + \gamma = 0 \]

Comparing coefficients:
\[ \begin{align} \alpha &= -2\sqrt{3} \\ \beta &= 3\sqrt{3} \\ \gamma &= 6 \end{align}

Step 5: Calculate \( \alpha^2 + \beta^2 + \gamma^2 \)

Compute each squared term:
\[ \begin{align} \alpha^2 &= (-2\sqrt{3})^2 = 12 \\ \beta^2 &= (3\sqrt{3})^2 = 27 \\ \gamma^2 &= 6^2 = 36 \end{align} \]

Sum them up:
\[ \alpha^2 + \beta^2 + \gamma^2 = 12 + 27 + 36 = 75 \]

Step 6: Match with Options

The correct answer corresponds to option (4).

Answer 2

Given, \[ x = 3\left(\frac{\tan\theta + \sqrt{3}}{1 - \sqrt{3}\tan\theta}\right) \] \[ x - \sqrt{3}\tan\theta = 3\tan\theta + 3\sqrt{3} \] \[ \tan\theta = \frac{x - 3\sqrt{3}}{3 + \sqrt{3}x} \quad \text{...(1)} \] Now, \[ 2\left(\frac{\tan\theta + \frac{1}{\sqrt{3}}}{1 - \frac{\tan\theta}{\sqrt{3}}}\right) = y \] \[ 2(\sqrt{3}\tan\theta + 1) = y(\sqrt{3} - \tan\theta) \quad \text{...(2)} \] Using (1) and (2), \[ 2\left(\frac{x - 3\sqrt{3}}{\sqrt{3}x + 3} + 1\right) = y\left(\sqrt{3} - \frac{x - 3\sqrt{3}}{\sqrt{3}(3 + x)}\right) \] Simplify: \[ 2\sqrt{3}(x - 3\sqrt{3} + x + \sqrt{3}) = y[3(\sqrt{3} + x) - x + 3\sqrt{3}] \] \[ 4\sqrt{3}x - 12 = y(2x + 6\sqrt{3}) \] \[ xy - 2\sqrt{3}x + 3\sqrt{3}y - 6 = 0 \] Comparing with general form \(xy + \alpha x + \beta y + \gamma = 0\), \[ \alpha = -2\sqrt{3}, \quad \beta = 3\sqrt{3}, \quad \gamma = -6 \] \[ \alpha^2 + \beta^2 + \gamma^2 = 12 + 27 + 36 = 75 \] \[ \boxed{\alpha^2 + \beta^2 + \gamma^2 = 75} \]