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  • Solve it, - The tangent to the curve,passing through the point (1,e) also passes through the point: - Limit , continuity and differentiability - JEE Main

The tangent to the curve, y = xe^{x^2} passing through the point (1,e) also passes through the point:

  • Option 1)

    (2,3e)

  • Option 2)

    (3,6e)

  • Option 3)

    \left (\frac{4}{3}, 2 \right )

  • Option 4)

    \left (\frac{5}{3}, 2 \right )

Answers (1)

best_answer

 

Equation of the tangent -

To find the equation of the tangent we need either one slope + one point or two points.

\therefore \:\:(y-y_{\circ})=m(x_{\circ }-y_{\circ })
 

or\:\:(y-y_{2})=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{2})

- wherein

Where  (x_{\circ},y_{\circ})   is the point on the curve and M = MT  slope of tangent.

Given curve,

y=xe^{x^{2}}

from the concept of tangent 

\frac{dy}{dx}_{(1,e)}=(e^{x^{2}}\cdot 2x+e^{x^{2}})_{(1,e)}=2e+e=3e

\therefore   Equation of tangent 

(y-e)=3e(x-1)

y=3e(x)-2e

(\frac{4}{3},2e)    lies  on it.

 

 


Option 1)

(2,3e)

Option 2)

(3,6e)

Option 3)

\left (\frac{4}{3}, 2 \right )

Option 4)

\left (\frac{5}{3}, 2 \right )

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