JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
The locus of the point of intersection of the lines \((\sqrt{3}) kx + ky -4 \sqrt{3}=0\) and \(\sqrt{3} x-y-4(\sqrt{3}) k=0\) is a conic, whose eccentricity is .............
- A \(0\)
- B \(2\)
- C \(4\)
- D \(8\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\(K =\frac{4 \sqrt{3}}{\sqrt{3} x + y }=\frac{\sqrt{3} x - y }{4 \sqrt{3}}\) \(\Rightarrow 3 x ^{2}- y ^{2}=48\) \(\Rightarrow \frac{ x ^{2}}{16}-\frac{ y ^{2}}{48}=1\) Now, \(48=16\left( e ^{2}-1\right)\) \(\Rightarrow e =\sqrt{4}=2\)
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