wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Find a point on parabola f(x)=(x3)2, where tangent is parallel to the chord joining the points (3,0) and (4,1)

Open in App
Solution

f(x)=(x3)2.....(1)(Given)
The given points are (3,0) and (4,1).
Therefore,
f(x)=(x3)2;x[3,4]
Since f(x) is a polynomial function, f(x) is continuous in [3,4].
f'(x)=2(x3) which exists in (3,4)
Therefore it is also differentiable in (3,4).
Hence both the conditions of Lagrange's Mean value theorem is satisfied.
Therefore,
f(3)=(33)2=0
f(4))=(43)2=1
f'(c)=2(c3)
Therefore,
2c6=1043
2c6=1
c=72
72(3,4)
Since c is the x-coordinate of that point at which the tangent is parallel to the chord joining the points (3,0) and (4,1).
Therefore Substituting x=72 in eqn(1), we have
f(x)=(723)2=14
Thus the required point is (72,14).
Hence the correct answer is (72,14).

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Line and a Parabola
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon