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  • Find the shortest distance between the circle <img alt="\mathrm{x^2+y^2-24 y+128=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2

Find the shortest distance between the circle \mathrm{x^2+y^2-24 y+128=0} and the parabola
\mathrm{y^2=4 x}

 

Option: 1

\mathrm{\sqrt{80}+4}


Option: 2

\mathrm{\sqrt{80}}


Option: 3

\mathrm{\sqrt{80}-4}


Option: 4

\mathrm{\sqrt{70}-4}


Answers (1)

best_answer

We know that shortest distance between two curves occurs along the common normal. Any normal to parabola will be of the form \mathrm{y+t x-2 t-t^3=3}. If it is common normal, of circle and parabola, it will passes through the centre of the circle.

\mathrm{\begin{array}{ll} \Rightarrow & 12=2 t+t^3 \\ \Rightarrow & (t-2)\left(t^2+2 t+6\right)=0 \\ \Rightarrow & t=2 \end{array}}

so point on parabola is (4, 4) distance of this point to centre of a circle will be \mathrm{\sqrt{80}} , hence shortest distance is \mathrm{\sqrt{80}-4=4(\sqrt{5}-1)}.

 

Posted by

Kuldeep Maurya

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