Question

$\text{The locus of the point of intersecion of the lines}\sqrt{3}x-y-4\sqrt{3}\gamma =0and\sqrt{3}\gamma x+\gamma y-4\sqrt{3}=0isahyperbolaofeccentricity$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B

2

$\text{The equation of lines}\sqrt{3}x-y-4\sqrt{3}\gamma =0and\sqrt{3}\gamma x+\gamma y-4\sqrt{3}=0canberewrittenas\sqrt{3}x-y=4\sqrt{3}\gamma and\sqrt{3}\gamma x+\gamma y=4\sqrt{3}respectively.$

$Multiplyingtheequations:$

$\sqrt{3}\gamma x^2-\gamma y^2=48\gamma$

$\Rightarrow \frac{3\gamma x^2}{48\gamma }-\frac{\gamma y^2}{48\gamma }=1\Rightarrow \frac{x^2}{16}-\frac{y^2}{48}=1$

$\text{This is the standard equation of a hyperbola where}{a}^2=16and{b}^2=48.$

$Eccebtricity,e=\sqrt{\frac{a^2+b^2}{a^2}}$

$\Rightarrow e=\sqrt{\frac{16+48}{16}}\Rightarrow e=\frac{8}{4}\Rightarrow e=2$