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  • Minimum distance between the curve  and <img alt="x^2+y^2-12 x+31=0" src="https://e

Minimum distance between the curve y^2=4 x and x^2+y^2-12 x+31=0 is .......

Option: 1

0


Option: 2

5


Option: 3

\sqrt{5}


Option: 4

2 \sqrt{5}


Answers (1)

best_answer

Centre and radius of the given circle is (6,0) and \sqrt{5}respectively.

Equation of normal for y^2=4 x \quad \text { at } \quad\left(t^2, 2 t\right) is y=-t x+2 t+t^3 , it must pass through (6,0) in order that it gives minimum distance between the two curves.

\begin{array}{ll} \\\therefore & 0=t^3-4 t \\ \\\Rightarrow & t=0 \text { or } t= \pm 2 \\ \\\therefore & A(4,4) \text { and } C(4,-4) \\ \\P A=P C=2 \sqrt{5} \end{array}

\therefore Required minimum distance 2 \sqrt{5} unit

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