A) 3
B) 9
C) ?9
D) ?3
Correct Answer: B
Solution :
Any normal is \[y+tx=6t+3{{t}^{3}}.\] It is identical with \[x+y=k\] if \[\frac{t}{1}=\frac{1}{1}=\frac{6t+3{{t}^{3}}}{k}\] \[\therefore \]\[t=1\] and \[(a,\,\,\pm \,2a).\] Þ \[k=9.\]You need to login to perform this action.
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