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  • Minimum distance between the curve and <img alt="\mathrm{x^{2}+y^{2}-12 x+3

Minimum distance between the curve \mathrm{y^{2}=4 x} and \mathrm{x^{2}+y^{2}-12 x+31=0}  is equal to

Option: 1

\sqrt{21}


Option: 2

\sqrt{26}-\sqrt{5}


Option: 3

\sqrt{21}-\sqrt{5}


Option: 4

\sqrt{28}-\sqrt{5}


Answers (1)

Shortest distance will take place along the common normal.


Equation of normal for \mathrm{y}^{2}=4 \mathrm{x}$ at $\left(\mathrm{a} \mathrm{t}^{2}, 2 \mathrm{t}\right) is

\mathrm{y=-t x+t+t^{3}}
If it passes through  \mathrm{(6,0)}, then
\mathrm{\mathrm{t}^{3}-5 \mathrm{t}=0 }
\mathrm{\Rightarrow \quad \mathrm{t}=0, \mathrm{t}= \pm \sqrt{5} }.
\mathrm{\Rightarrow \quad \mathrm{A} \equiv(5,2 \sqrt{5}), \mathrm{C} \equiv(5,-2 \sqrt{5}) }.

\mathrm{Now,\mathrm{PA}=\mathrm{PC}=\sqrt{1+20}=\sqrt{21}, \mathrm{OP}=6 }

\mathrm{\therefore \text{Minimum distance =} (\sqrt{21}-\sqrt{5})\: units}.
 

Posted by

Ramraj Saini

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