Question

The points on the hyperbola $x^2-y^2=2$ closest to the point (0, 1) are

• A. $(\pm \frac{3}{2},\frac{1}{2})$
• B. $(\frac{1}{2},\pm \frac{3}{2})$
• C. $(\frac{1}{2},\frac{1}{2})$
• D. $(\pm \frac{3}{4}\pm \frac{3}{2})$

Answer 1

The correct option is A $(\pm \frac{3}{2},\frac{1}{2})$

Let point be $(\sqrt{2}sec\theta ,\sqrt{2}tan\theta )$ closest to (0, 1)
So distance $=\sqrt{2se{c}^2\theta +(\sqrt{2}tan\theta -1{)}^2}$
$=\sqrt{2se{c}^2\theta +2ta{n}^2\theta -2\sqrt{2}tan\theta +1}$
$=\sqrt{4ta{n}^2\theta -2\sqrt{2}tan\theta +3}$
$=2\sqrt{ta{n}^2\theta -\frac{tan\theta }{\sqrt{2}}+\frac{3}{4}}$
$=2\sqrt{(tan\theta -\frac{1}{2\sqrt{2}}{)}^2+\frac{3}{4}-\frac{1}{8}}$
So distance is min when $tan\theta =\frac{1}{2\sqrt{2}}$
$sec\theta =\pm \frac{3}{2\sqrt{2}}$