A) A hyperbola with length of its transeverse axis\[8\sqrt{2}\]
B) An ellipse with length of its major axis \[8\sqrt{2}\]
C) An ellipse whose eccentricity is \[\frac{1}{\sqrt{3}}\]
D) A hyperbola whose eccentricity is \[\sqrt{3}\]
Correct Answer: A
Solution :
Gives lines are: \[\sqrt{2x}-y+4\sqrt{2k}=0\] \[\Rightarrow \]\[\sqrt{2x}+4\sqrt{2k}=y\] ??(i) and \[\sqrt{2kx}+ky-4\sqrt{2}=0\]?.. (ii) We have from the equations of the lines: Substituting (i) in (ii), \[\Rightarrow 2\sqrt{2}kx+4\sqrt{2}({{k}^{2}}-1)=0\] \[\Rightarrow \]\[x=\frac{1(1-{{k}^{2}})}{k},y=\frac{2\sqrt{2}(1+{{k}^{2}})}{k}\] \[\Rightarrow \]\[{{\left( \frac{y}{4\sqrt{2}} \right)}^{2}}-{{\left( \frac{x}{4} \right)}^{2}}=1\] \[\Rightarrow \]\[{{\left( \frac{y}{4\sqrt{2}} \right)}^{2}}-{{\left( \frac{x}{4} \right)}^{2}}=1\] Locus of transverse axis \[=2\sqrt{32}\] \[=2\times 4\sqrt{2}\] \[=8\sqrt{2}\] Thus, the locus is a hyperbola with length of its transverse axis equal to\[8\sqrt{2}.\] So option A is the correct answer.
You need to login to perform this action.
You will be redirected in
3 sec
