A) \[yt=x+a{{t}^{2}}\]
B) \[y=xt+a{{t}^{2}}\]
C) \[y=xt+\frac{a}{t}\]
D) \[y=tx\]
Correct Answer: A
Solution :
Any point on \[{{y}^{2}}=4ax\]is\[(a{{t}^{2}},\,2at)\], then tangent is\[2aty=2a(x+{{t}^{2}})\]Þ \[yt=x+a{{t}^{2}}\].You need to login to perform this action.
You will be redirected in
3 sec