A) \[\frac{3}{4}\]
B) \[\frac{4}{3}\]
C) 2
D) 3
Correct Answer: B
Solution :
Given equation of conic \[3{{x}^{2}}-4y+6x-3=0\] \[\Rightarrow \] \[3{{x}^{2}}+6x-3=4y\] \[\Rightarrow \] \[3({{x}^{2}}+2x-1)=4y\] \[\Rightarrow \] \[3({{x}^{2}}+2x+1-2)=4y\] \[\Rightarrow \] \[3{{(x+1)}^{2}}-6=4y\] \[\Rightarrow \] \[3{{(x+1)}^{2}}=4y+6\] \[\Rightarrow \] \[{{(x+1)}^{2}}=\frac{4}{3}(y+3/2)\] ??(i) Let \[{{X}^{2}}=\frac{4}{3}Y\] ...(ii) where \[X=x+1\] and \[Y=y+3/2\] and \[4b=4/3\Rightarrow \,\,b=1/3\] So, now the length of latuserectum is 4/3.
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