• question_answer The tangent to the curve, $y=x{{e}^{{{x}^{2}}}}$ passing, through the point $(1,e)$ also passes through the point: A) $(2,3e)$ B) $\left( \frac{4}{3},2e \right)$ C) $\left( \frac{5}{3},2e \right)$ D) $(3,6e)$

 $y=x{{e}^{{{x}^{2}}}}$ $(1,e)$lies on this Now, $\frac{dy}{dx}=x{{e}^{{{x}^{2}}}}.2x+{{e}^{{{x}^{2}}}}.1$ Put $x=1$ $m=2e+e=3e$ Equation of tangent at $(1,e)$ $y-e=3e(x-1)$ $y-e=3ex-3e$ $y=3ex-2e$ $\left( \frac{4}{3},2e \right)$satisfies it $\therefore$Answer is B.