Straight Line and Pair of Straight Lines 1 Question 13

13. The tangent to the curve, y=xex2 passing through the point (1,e) also passes through the point

(2019 Main, 10 Jan II)

(a) 43,2e

(b) (3,6e)

(c) (2,3e)

(d) 53,2e

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Solution:

  1. Given equation of curve is y=xex2

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

Now, slope of tangent, at point (1,e), to the curve (i) is

dydx|(1,e)=x(2x)ex2+ex2=2e+e=3e

Now, equation of tangent is given by

(yy1)=m(xx1)ye=3e(x1)y=3ex2e

On checking all the options, the option 43,2e satisfy the equation of tangent.



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