Q. If $x = 8 + 3\sqrt{7}$ and $xy = 1$, then the value of $\frac{1}{x^{2}} + \frac{1}{y^{2}}$ is

A) 254
B) 192
C) 292
D) 66
Explanation

(a)
$\because\ \ \ x = 8 + 3\sqrt{7}$
$\therefore y = \frac{1}{8 + 3\sqrt{7}} = 8 - 3\sqrt{7}$
Now, $\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{x^{2} + y^{2}}{(xy)^{2}}$
$= (x + y)^{2} - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\because xy = 1\rbrack$
$= \left( 8 + 3\sqrt{7} + 8 - 3\sqrt{7} \right)^{2} - 2$
$= (16)^{2} - 2 = 254$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.