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Minimum distance between the parabolas \mathrm{y}^{2}-4 \mathrm{x}-8 \mathrm{y}+40=0 and \mathrm{\mathrm{x}^{2}-8 \mathrm{x}-4 \mathrm{y}+40=0} is

Option: 1

0


Option: 2

\sqrt{3}


Option: 3

2 \sqrt{2}


Option: 4

\sqrt{2}


Answers (1)

best_answer

Since two parabolas are symmetrical about \mathrm{y}=\mathrm{x}
Minimum distance is distance between tangents to the parabola parallel to \mathrm{y}=\mathrm{x}.

Differentiating \mathrm{x}^{2}-8 \mathrm{x}-4 \mathrm{y}+40=0$ w.r.t $\mathrm{x}, we get \mathrm{2 \mathrm{x}-8-4 \mathrm{y}^{\prime}=0}

\mathrm{\mathrm{y}^{\prime}=\frac{\mathrm{x}-4}{2}=1}
\mathrm{x=6\, and \, y=7}
 Corresponding point on \mathrm{(\mathrm{y}-4)^{2}=4(\mathrm{x}-6)} is (7,6) so minimum distance =\sqrt{2}.

Posted by

Pankaj Sanodiya

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