JEE PYQ: Hyperbola Question 4
Question 4 - 2021 (25 Feb Shift 1)
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 ______.
Show Answer
Answer: (2)
Solution
From the lines: $k = \frac{4\sqrt{3}}{\sqrt{3}x+y}$ and $k = \frac{\sqrt{3}x - y}{4\sqrt{3}}$. Eliminating $k$: $\frac{(\sqrt{3}x - y)(\sqrt{3}x + y)}{4\sqrt{3}} = 4\sqrt{3}$. So $3x^2 - y^2 = 48$, giving $\frac{x^2}{16} - \frac{y^2}{48} = 1$. Eccentricity $= \sqrt{1 + \frac{48}{16}} = \sqrt{4} = 2$.
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