If for , the points lie on , then is equal to [2025]

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Solution

Q.
If for $θ \in [- \frac{π}{3}, 0]$ , the points $(x, y) = (3 \tan (θ + \frac{π}{3}), 2 \tan (θ + \frac{π}{6}))$ lie on $x y + α x + β y + γ = 0$ , then $α^{2} + β^{2} + γ^{2}$ is equal to [2025]

1 72

2 75

3 80

4 96

Ans.
(2)

We have, $x = 3 \tan (θ + \frac{π}{3})$

$\Rightarrow x = 3 (\frac{\tan θ + \sqrt{3}}{1 - \sqrt{3} \tan θ})$

$\Rightarrow x - x \sqrt{3} \tan θ = 3 \tan θ + 3 \sqrt{3}$

$\Rightarrow \tan θ = \frac{x - 3 \sqrt{3}}{3 + \sqrt{3} x}$ ... (i)

Also, $y = 2 \tan (θ + \frac{π}{6})$

$\Rightarrow y = 2 \frac{(\tan θ + \frac{1}{\sqrt{3}})}{(1 - \frac{\tan θ}{\sqrt{3}})}$

$\Rightarrow y (\sqrt{3} - \tan θ) = 2 (\sqrt{3} \tan θ + 1)$ ... (ii)

Solving (i) and (ii), we get

$\Rightarrow 2 (\frac{x - 3 \sqrt{3}}{\sqrt{3} + x} + 1) = y (\sqrt{3} - \frac{(x - 3 \sqrt{3})}{\sqrt{3} (\sqrt{3} + x)})$

$\Rightarrow 2 \sqrt{3} (x - 3 \sqrt{3} + x + \sqrt{3}) = y (3 (\sqrt{3} + x) - x + 3 \sqrt{3})$

$\Rightarrow 4 \sqrt{3} x - 12 = y (2 x + 6 \sqrt{3})$

$\Rightarrow x y - 2 \sqrt{3} x + 3 \sqrt{3} y + 6 = 0$

Comparing with $x y + α x + β y + γ$ , we get

$α = - 2 \sqrt{3}, β = 3 \sqrt{3}, γ = 6$

$∴ α^{2} + β^{2} + γ^{2} = 12 + 27 + 36 = 75$ .

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