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  • I have a doubt, kindly clarify. - The shortest distance between the point and the curve, is: - Limit , continuity and differentiability - JEE Main

The shortest distance between the point(3/2,0) and the curve y=\sqrt{x}, (x>0) is:

  • Option 1)

     

    \sqrt{5}/2

  • Option 2)

     

    \sqrt{3}/2

  • Option 3)

     

    3/2

  • Option 4)

     

    5/4

Answers (1)

best_answer

 

Method for maxima or minima -

By second derivative method :

Step\:1.\:\:find\:values\:of\:x\:for\:\frac{dy}{dx}=0

Step\:\:2.\:\:\:x=x_{\circ }\:\:is\:a\:point\:of\:local\:maximum\:if  f''(x)<0\:\:and\:local\:minimum\:if\:f''(x)>0

- wherein

Where\:\:y=f(x)

\frac{dy}{dx}=f'(x)

 

 038???

Given curve is

y2 = x

Let point on the curve is ( t2 , t )

Distance between \left ( \frac{3}{2} , 0 \right ) and ( t2 , t )  is

D = \sqrt{\left ( t^{2}-\frac{3}{2} \right )^{2} + t^{2}}

= \sqrt{ t^{4}-2t ^{2} + \frac{9}{4}} = \sqrt{\left ( t^{2} - 1 \right )^{2} + \frac{5}{4}}

So minimum distance when t = 1 or -1 which is \frac{\sqrt{5}}{2} .


Option 1)

 

\sqrt{5}/2

Option 2)

 

\sqrt{3}/2

Option 3)

 

3/2

Option 4)

 

5/4

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