Tough JEE Question? Convert Parametric tan–Form to Line Equation FAST ⚡

 

❓ Question

FOR:

If for

$$\theta \in [-\frac{\pi }{3},0],(x,y)=(3\tan \text{\hspace{0.17em}⁣}(\theta +\frac{\pi }{3}),2\tan \text{\hspace{0.17em}⁣}(\theta +\frac{\pi }{6}))$$

lie on the curve

$$xy + \alpha x + \beta y + \gamma = 0,$$

then the value of

$$\alpha^2 + \beta^2 + \gamma^2$$

is equal to ?

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🖼️ Question Image

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✍️ Short Solution

We are told that for every $\theta$ in the interval,

$$x = 3\tan\left(\theta + \frac{\pi}{3}\right), \quad y = 2\tan\left(\theta + \frac{\pi}{6}\right)$$

always satisfies

$$xy + \alpha x + \beta y + \gamma = 0.$$

That means there is a fixed relation between $x$ and $y$ which does not involve $\theta$. We’ll eliminate $\theta$ using a standard tangent identity.

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🔹 Step 1 — Define new angles and variables

Let

$$A = \theta + \frac{\pi}{3}, \quad B = \theta + \frac{\pi}{6}.$$

Then,

$$x = 3\tan A,\quad y = 2\tan B.$$

Compute the difference:

$$A - B = \left(\theta + \frac{\pi}{3}\right) - \left(\theta + \frac{\pi}{6}\right) = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}.$$

Let

$$u = \tan A,\quad v = \tan B.$$

So

$$x=3u,y=2v\mathrm{.}$$

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🔹 Step 2 — Apply $\tan(A - B)$ identity

We know

$$\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}=\tan \left(\frac{\pi }{6}\right)=\frac{1}{\sqrt{3}}\mathrm{.}$$

So,

$$\frac{u-v}{1+uv}=\frac{1}{\sqrt{3}}\mathrm{.}$$

Cross-multiply:

$$\sqrt{3}(u-v)=1+uv\mathrm{.}$$

Rearrange to get a single equation:

$$uv-\sqrt{3}u+\sqrt{3}v+1=0.$$

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🔹 Step 3 — Convert back to x and y

Recall

$$u = \frac{x}{3},\quad v = \frac{y}{2}.$$

Then

$$uv = \frac{x}{3}\cdot\frac{y}{2} = \frac{xy}{6}.$$

Substitute:

$$\frac{xy}{6}-\sqrt{3}\left(\frac{x}{3}\right)+\sqrt{3}\left(\frac{y}{2}\right)+1=0.$$

Simplify coefficients:

$$\frac{xy}{6}-\frac{\sqrt{3}}{3}x+\frac{\sqrt{3}}{2}y+1=0.$$

Multiply throughout by 6:

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

Compare with

$$xy + \alpha x + \beta y + \gamma = 0,$$

we get

$$\alpha =-2\sqrt{3},\beta =3\sqrt{3},\gamma =6.$$

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🔹 Step 4 — Compute $\alpha^2 + \beta^2 + \gamma^2$

$${\alpha }^2=(-2\sqrt{3}{)}^2=4\cdot 3=12,$$$${\beta }^2=(3\sqrt{3}{)}^2=9\cdot 3=27,$$$${\gamma }^2=6^2=36.$$

So,

$${\alpha }^2+{\beta }^2+{\gamma }^2=12+27+36=75.$$

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✅ Final Answer

$$\boxed{75}$$

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