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Solve it, - Co-ordinate geometry - JEE Main-8

The shortest distance between the line y=x and the curve y^{2}=x-2 is :
 

 

  • Option 1)

    2

  • Option 2)

    \frac{7}{8}

  • Option 3)

    \frac{11}{4\sqrt{2}}

     

  • Option 4)

    \frac{7}{4\sqrt{2}}

 
Answers (1)
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\\Line\; \; y=x\\\; \; \; \\Curve=y^{2}=x-2

Find shortest distance

 

\\y^{2}=x-2\\\; \; \; \\2y.y{'}=1

y'=\frac{1}{2y}

the tangent at point P is parallel to x=y

So y'=1=\frac{1}{2y}

=y=\frac{1}{2}

putiing the value of y in curve

=\left ( \frac{1}{2} \right )^{2}=x-2\Rightarrow \frac{1}{4}=x-2

                                             x=\frac{1}{4}+2=\frac{9}{4}

\\P=\left ( \frac{9}{4},\frac{1}{2} \right )

Shortest distance between two parallel lines or perpendicular distance from P\; to\; x=y

=\left | \frac{\frac{9}{4}-\frac{1}{2}}{\sqrt{1^{2}+1^{2}}} \right |=\frac{\frac{7}{4}}{\sqrt{2}}=\frac{7}{4\sqrt{2}}

 

 

 


Option 1)

2

Option 2)

\frac{7}{8}

Option 3)

\frac{11}{4\sqrt{2}}

 

Option 4)

\frac{7}{4\sqrt{2}}

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