Question

Let a, b, c be positive real numbers. The following system of equations in x, y, z:-

$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1, \frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$

has:-

• A. no solution
• B. unique solution
• C. infinitely many solutions
• D. finitely many solutions

Answer 1

Answer: (C)

infinitely many solutions

Answer 2

The given system of equations is:

1. $\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$
2. $\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{z^2}{c^2}=1$

Subtract equation (2) from equation (1):

$$\left(\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}\right) - \left(\frac{x^2}{a^2}-\frac{y^2}{b^2}+\frac{z^2}{c^2}\right) = 1 - 1$$ $$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} - \frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2} = 0$$ $$2\frac{y^2}{b^2} = 0$$

Since $b$ is a positive real number, $b^2 \ne 0$. Therefore, we must have:

$$y^2 = 0 \implies y = 0$$

Now, substitute $y=0$ back into equation (1):

$$\frac{x^2}{a^2}+\frac{0^2}{b^2}+\frac{z^2}{c^2}=1$$ $$\frac{x^2}{a^2}+\frac{z^2}{c^2}=1$$

This is the equation of an ellipse in the xz-plane. Since $a$ and $c$ are positive real numbers, this ellipse is a well-defined curve. An ellipse contains infinitely many points. For every point $(x, z)$ on this ellipse, the triplet $(x, 0, z)$ is a solution to the given system of equations.

Since there are infinitely many points on an ellipse, the system has infinitely many solutions.