Q. Let $a,\ b,\ c$ be positive numbers. The following system of equations in $x,\ y\ and\mathcal{\ z,}\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{\mathcal{z}^{2}}{c^{2}} = 1;\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{\mathcal{z}^{2}}{c^{2}} = 1$ and $\frac{- x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{\mathcal{z}^{2}}{c^{2}} = 1$ has

A) No solution
B) Unique solution
C) Infinitely many solutions
D) Finitely may solutions
Explanation

(d)
Given system of equation is
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{\mathcal{z}^{2}}{c^{2}} = 1$ ...(i)
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} + \frac{\mathcal{z}^{2}}{c^{2}} = 1$ ...(ii)
and $–\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{\mathcal{z}^{2}}{c^{2}} = 1$ ...(iii)
On adding all these equations, we get
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{\mathcal{z}^{2}}{c^{2}} = 3$ …(iv)
On subtracting Eq. (i) from Eq. (iv), Eq. (ii) from Eq. (iv) and Eq. (iii) from Eq. (iv),we get
$\frac{{2\mathcal{z}}^{2}}{c^{2}} = 2,\ \frac{{2y}^{2}}{b^{2}}$=2, $\frac{{2x}^{2}}{a^{2}} = 2$
$\mathcal{\Rightarrow z = \pm}c,\ y = \pm b,\ x = \pm a$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

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