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

Let's simplify the notation by introducing new variables: Let $X = \frac{x^2}{a^2}$, $Y = \frac{y^2}{b^2}$, and $Z = \frac{z^2}{c^2}$. Since $a, b, c$ are positive real numbers, $a^2, b^2, c^2$ are positive. Also, $x^2, y^2, z^2$ are non-negative. Therefore, $X \ge 0$, $Y \ge 0$, and $Z \ge 0$.

The system of equations can be rewritten as:

1. $X + Y + Z = 1$
2. $-X + Y + Z = 1$

Now, we solve this system of linear equations for $X, Y, Z$:

Step 1: Add Equation (1) and Equation (2) $(X + Y + Z) + (-X + Y + Z) = 1 + 1$ $2Y + 2Z = 2$ $Y + Z = 1$

Step 2: Subtract Equation (2) from Equation (1) $(X + Y + Z) - (-X + Y + Z) = 1 - 1$ $X + Y + Z + X - Y - Z = 0$ $2X = 0$ $X = 0$

Step 3: Substitute back the original expressions for X, Y, Z From $X = 0$: $\frac{x^2}{a^2} = 0 \implies x^2 = 0 \implies x = 0$

From $Y + Z = 1$: $\frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$

So, the original system of equations simplifies to: $x = 0$ $\frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$

Step 4: Analyze the equation $\frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$ This is the equation of an ellipse centered at the origin in the y-z plane. Since $y$ and $z$ are real variables, we need to find how many real pairs $(y, z)$ satisfy this equation.

We can express $z^2$ in terms of $y^2$: $\frac{z^2}{c^2} = 1 - \frac{y^2}{b^2}$ $z^2 = c^2 \left(1 - \frac{y^2}{b^2}\right)$

For $z$ to be a real number, $z^2$ must be non-negative. Since $c^2 > 0$ (as $c$ is a positive real number), we must have: $1 - \frac{y^2}{b^2} \ge 0$ $\frac{y^2}{b^2} \le 1$ $y^2 \le b^2$ This implies $-b \le y \le b$.

For any real value of $y$ in the interval $[-b, b]$:

• If $y \in (-b, b)$, then $1 - \frac{y^2}{b^2} > 0$, which means $z = \pm c \sqrt{1 - \frac{y^2}{b^2}}$. This gives two distinct real values for $z$. Since there are infinitely many real numbers in the interval $(-b, b)$, there are infinitely many possible values for $y$, each yielding two $z$ values.
• If $y = b$ or $y = -b$, then $1 - \frac{y^2}{b^2} = 0$, which means $z = 0$. This gives one real value for $z$.

Since there are infinitely many real numbers in the interval $[-b, b]$, there are infinitely many possible values for $y$. For each such $y$, there is at least one corresponding real value for $z$. Therefore, there are infinitely many pairs $(y, z)$ that satisfy the equation $\frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$.

Since $x=0$ is a fixed value, and there are infinitely many pairs $(y, z)$ satisfying the condition, the system has infinitely many solutions $(x, y, z)$.

For example, some solutions are:

• $(0, 0, c)$
• $(0, 0, -c)$
• $(0, b, 0)$
• $(0, -b, 0)$
• $(0, b/2, \frac{\sqrt{3}}{2}c)$
• $(0, b/2, -\frac{\sqrt{3}}{2}c)$ And so on.