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Let \mathrm{P} and \mathrm{Q}  be any points on the curves \mathrm{(x-1)^{2}+(y+1)^{2}=1 \text { and } y=x^{2} \text {, }}  respectively. The distance between \mathrm{P} and \mathrm{Q}  is minimum for some value of the abscissa of \mathrm{P} in the interval

Option: 1

\mathrm{\left(0, \frac{1}{4}\right)}


Option: 2

\mathrm{\left(\frac{1}{2}, \frac{3}{4}\right)}


Option: 3

\mathrm{\left(\frac{1}{4}, \frac{1}{2}\right)}


Option: 4

\mathrm{\left(\frac{3}{4}, 1\right)}


Answers (1)

best_answer

For maximum distance, normal at \mathrm{P(t,t^{2})}  should also be normal at \mathrm{Q\Rightarrow } normal at \mathrm{P } should pass through (1,-1)

For equation of normal at \mathrm{P}

\mathrm{y=x^{2}} \\

\mathrm{\frac{d y}{d x}=2 x} \\

\mathrm{m_{N}=-\frac{1}{2 x}=-\frac{1}{2 t}}

\mathrm{Equation: y-t^{2}=-\frac{1}{2 t}(x-t)}

\mathrm{\Rightarrow 2 y t-2 t^{3}=-x+t }\\

\mathrm{\Rightarrow x+2 t y-t-2 t^{3}=0}

\mathrm{It\; passes \;through (1 ;-1)}

\mathrm{1-2 t-t-2 t^{3}=0} \\

\mathrm{\Rightarrow 2 t^{3}+3 t-1=0} \\

\mathrm{f\left(\frac{1}{4}\right)=\frac{2}{64}+\frac{3}{4}-1<0} \\

\mathrm{f\left(\frac{1}{2}\right)=\frac{2}{8}+\frac{3}{2}-1>0} \\

\mathrm{\therefore \text { Root of } 2 t^{3}+3 t-1 \text { lies } \text { in interval }\left(\frac{1}{4}, \frac{1}{2}\right)} \\

Hence correct option is 3

 

 

Posted by

Ajit Kumar Dubey

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