WBJEE · Maths · Straight Lines
For different values of \(\alpha\), the locus of the point of intersection of the two straight lines \(\sqrt{3} x-y-4 \sqrt{3} \alpha=0\) and \(\sqrt{3} \alpha x+\alpha y-4 \sqrt{3}=0\) is
- A a hyperbola with eccentricity 2
- B an ellipse with eccentricity \(\sqrt{\frac{2}{3}}\)
- C a hyperbola with eccentricity \(\sqrt{\frac{19}{16}}\)
- D an ellipse with eccentricity \(\frac{3}{4}\)
Answer & Solution
Correct Answer
(A) a hyperbola with eccentricity 2
Step-by-step Solution
Detailed explanation
Hints : \(\sqrt{3} x-y=4 \sqrt{3} \alpha\) (1) \(\sqrt{3} x+y=\frac{4 \sqrt{3}}{\alpha}\) (1) \(x(2) \Rightarrow 3 x^2-y^2=48 \Rightarrow \frac{x^2}{16}-\frac{y^2}{48}=1\) \[ \mathrm{e}=\sqrt{\frac{48+16}{16}}=2 \]
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