TestHub
TestHub

Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeMaxima-MinimaMedium2 minPYQ_2022
MathematicsMediumsingle choice

LetPandQbe any points on the curvesx-12+y+12=1andy=x2, respectively. The distance betweenPandQis minimum for some value of the abscissa ofPin the interval

Options:

Answer:
C
Solution:

Let Px1,x12

Minimum distance will be obtained at common normal of the parabola and circle.

So for minimum PQ, the distance between centre of circle and P should be minimum.

Now, the distance of P from given circle,

d=x1-12+x12+12-1

For least value of d, we need to minimize

fx1=x1-12+x12+12

i.e. f'x1=2x1-1+4x1x12+1=0

From options

f'14 is -ve and f'12 is +ve

So, f'x1=0 for some x114,12 from IMVT

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Maxima-Minima
2mℹ️ Source: PYQ_2022

Doubts & Discussion

Loading discussions...