A) \[y-tx=a{{t}^{3}}+2at\]
B) \[y+tx=a{{t}^{3}}+2at\]
C) \[y-tx=a{{t}^{3}}-2at\]
D) None of these
Correct Answer: B
Solution :
The equation of normal to\[{{y}^{2}}=4\text{ }ax\]at point \[(a{{t}^{2}},2at)\]is \[y-2at=-\frac{2at}{2a}(x-a{{t}^{2}})\] \[\because \] \[\frac{dy}{dx}=\frac{2a}{2y}\] \[\Rightarrow \] \[\frac{-dx}{d{{y}_{(a{{t}^{2}},2at)}}}=\frac{-2at}{2a}\] \[\Rightarrow \] \[2ay-4{{a}^{2}}t=-2atx+2{{a}^{2}}{{t}^{3}}\] \[\Rightarrow \] \[y-2at=-tx+a{{t}^{3}}\] \[\Rightarrow \] \[y+tx=a{{t}^{3}}+2at\]
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