A) \[\frac{1}{t}\]
B) t
C) \[-t\]
D) \[{{t}^{2}}\]
Correct Answer: C
Solution :
[c] \[\because \]Tangent of the paralbola \[{{y}^{2}}\] \[=4ax\,be\,y.{{y}_{1}}=2a(x+{{x}_{1}})\] At\[\left( a{{t}^{2}},2at \right)\], tangent be \[y.2at=2a\left( x+a{{t}^{2}} \right)\] \[yt=\left( x+a{{t}^{2}} \right)\] \[y=\frac{1}{t}\left( x+at \right)\]slope of the tangent \[=\frac{1}{t}\] So, the slope of the Normal \[=-t\] Hence, the option [c] is correct.You need to login to perform this action.
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