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Step-by-Step Solution
Step 1: Express the given lines
The two lines given are:
1) \sqrt{3} \, kx + k \, y - 4\sqrt{3} = 0 \quad \dots (1)
2) \sqrt{3} \, x - y - 4(\sqrt{3})\, k = 0 \quad \dots (2)
Step 2: Rewrite and add/subtract the equations
First, rearrange equation (2) to have the same form in terms of k :
\sqrt{3} \, x - y = 4\sqrt{3}\, k
Multiply both sides by k to get:
\sqrt{3} \, kx - k \, y = 4\sqrt{3}\, k^2 \quad \dots (2')
So now we have from (1) and (2'):
(1) \sqrt{3} \, kx + k \, y = 4\sqrt{3}
(2') \sqrt{3} \, kx - k \, y = 4\sqrt{3}\, k^2
Add (1) and (2'):
(\sqrt{3} \, kx + k \, y) + (\sqrt{3}\, kx - k \, y) = 4\sqrt{3} + 4\sqrt{3}\, k^2
This simplifies to:
2\sqrt{3}\, kx = 4\sqrt{3}\, (k^2 + 1)
Divide both sides by 2\sqrt{3}\, k to obtain:
x = 2\left(k + \frac{1}{k}\right) \quad \dots (3)
Next, subtract (2') from (1):
(\sqrt{3}\, kx + k \, y) - (\sqrt{3}\, kx - k \, y) = 4\sqrt{3} - 4\sqrt{3}\, k^2
This simplifies to:
2k \, y = 4\sqrt{3}\,(1 - k^2)
Rewriting carefully (or from the original approach):
y = 2\sqrt{3}\left(\frac{1}{k} - k\right) \quad \dots (4)
Step 3: Eliminate the parameter k to find the locus
From (3): x = 2 \left( k + \frac{1}{k} \right).
We can write k + \frac{1}{k} = \frac{x}{2}.
From (4): y = 2\sqrt{3}\left(\frac{1}{k} - k\right).
We can write \frac{1}{k} - k = \frac{y}{2\sqrt{3}}.
Notice:
\left(k + \frac{1}{k}\right)^2 - \left(\frac{1}{k} - k\right)^2 = 4\,k\,\frac{1}{k} = 4.
Hence, substituting k + \frac{1}{k} = \frac{x}{2} and \frac{1}{k} - k = \frac{y}{2\sqrt{3}} :
\left(\frac{x}{2}\right)^2 - \left(\frac{y}{2\sqrt{3}}\right)^2 = 4.
This becomes:
\frac{x^2}{4} - \frac{y^2}{12} = 4.
Divide through by 4:
\frac{x^2}{16} - \frac{y^2}{48} = 1.
This is the standard form of a hyperbola:
\frac{(x^2)}{a^2} - \frac{(y^2)}{b^2} = 1,
where a^2 = 16 and b^2 = 48.
Step 4: Find the eccentricity of the hyperbola
For a hyperbola of form \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1, the eccentricity e satisfies
e^2 = 1 + \frac{b^2}{a^2}.
Here:
a^2 = 16, \quad b^2 = 48.
So,
e^2 = 1 + \frac{48}{16} = 1 + 3 = 4.
Therefore,
e = 2.
Answer: The eccentricity of the conic is 2.
