A) (3,-4), (13, 4)
B) (-3,-4), (13,-4)
C) (3, 4), (-13, 4)
D) (5,-8), (-5, 8)
Correct Answer: C
Solution :
Given that \[{{x}^{2}}+10x-16y+25=0\] \[\Rightarrow \,\,{{(x+5)}^{2}}=16\,y\] \[\Rightarrow \] \[{{X}^{2}}=4\,AY\] where \[X=x+5,\,A=4,\,Y=y\]. The ends of die latus rectum are \[\left( +2A,A \right)\] and \[\left( -2A,A \right)\] \[\Rightarrow \] \[x+5=2\,(4)\Rightarrow x=-8-5=3,\,y=4\] and \[x+5=-2\,(4)\Rightarrow x=-8-5=-13,\,y=4\] \[\Rightarrow \] (3, 4) and (-13, 4).You need to login to perform this action.
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