Question

If a, b, c $\in R^{+}$, then $\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=1$, $\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ and $-\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$ has?

• A. A unique solution
• B. No solution
• C. Infinitely many solutions
• D. Finitely many solutions

Answer 1

The correct option is A A unique solution

$\Delta =\left|\begin{array}{ccc}\frac{1}{a^2}& \frac{1}{b^2}& \frac{-1}{c^2}\\ \frac{1}{a^2}& \frac{-1}{b^2}& \frac{1}{c^2}\\ \frac{-1}{a^2}& \frac{1}{b^2}& \frac{1}{c^2}\end{array}\right|$

$=\frac{1}{a^2{b}^2{c}^2}\left|\begin{array}{ccc}1& 1& -1\\ 1& -1& 1\\ -1& 1& 1\end{array}\right|$

$=\frac{1}{a^2{b}^2{c}^2}(-4)\ne 0$

$\therefore$ Given system has uniques solution.