A) 5:4
B) 14:13
C) 2:1
D) 13:11
Correct Answer: A
Solution :
Equation of tangents \[{{y}^{2}}=12x\Rightarrow y=2x+\frac{3}{m}\] \[\frac{{{x}^{2}}}{1}-\frac{{{y}^{2}}}{8}=1\Rightarrow y=mx\pm \sqrt{{{m}^{2}}-8}\] Since they are common tangent \[\therefore \]\[\frac{3}{m}=\pm \sqrt{{{m}^{2}}-8}\] \[\frac{{{x}^{2}}}{1}-\frac{{{y}^{2}}}{8}=1\] \[{{m}^{4}}-8{{m}^{2}}-9=0\] \[e=3\] \[m=\pm 3\] \[ae=3\]You need to login to perform this action.
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