Question

The tangent to the curve, \(y = x{{\rm{e}}^{{x^2}}}\) through the point (1, e) also passes through the point:
1. (2, 3e)
2. \(\left( {\frac{4}{3},2{\rm{e}}} \right)\)
3. \(\left( {\frac{5}{3},2{\rm{e}}} \right)\)
4. (3, 6e)

Answer 1

Correct Answer - Option 2 : \(\left( {\frac{4}{3},2{\rm{e}}} \right)\)

From question, the equation of the curve given is:

\(y = x{{\rm{e}}^{{x^2}}}\)

Since, (1, e) lie on the curve, so the point of contact is (1, e).

Now, the slope of the tangent at the point (1, e) in the given curve is:

\(\Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1,e} \right)}} = {\left( {x\left( {2x} \right){e^{{x^2}}} + {e^{{x^2}}}} \right)_{\left( {1,e} \right)}}\)

\(\Rightarrow {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1,e} \right)}} = 2e + e\)

\(\therefore {\left( {\frac{{dy}}{{dx}}} \right)_{\left( {1,e} \right)}} = 3e\)

Now, the equation of tangent is given by the formula:

(y - y₁) = m(x - x₁)

On substituting the values,

⇒ y - e = 3e(x - 1)

∴ y = 3ex - 2e

From the points given in the option \(\left( {\frac{4}{3},\:2e} \right)\).

The point in option (b) satisfies the obtained equation. So, option (b) is the correct answer.