Find the nature of the quadric surface given by the equation $10x^2 + 5y^2 - 4z^2 - 60x + 10y + 75 = 0$.

Steps to Solve

1. Group like terms

Rearrange the given equation to group the $x$, $y$, and $z$ terms together:

$10x^2 - 60x + 5y^2 + 10y - 4z^2 + 75 = 0$

2. Complete the square for $x$ terms

To complete the square for $x$, we need to add and subtract $(60/(2*10))^2 * 10 = 90$ inside the equation:

$10(x^2 - 6x) + 5y^2 + 10y - 4z^2 + 75 = 0$

$10(x^2 - 6x + 9 - 9) + 5y^2 + 10y - 4z^2 + 75 = 0$

$10((x - 3)^2 - 9) + 5y^2 + 10y - 4z^2 + 75 = 0$

$10(x - 3)^2 - 90 + 5y^2 + 10y - 4z^2 + 75 = 0$

$10(x - 3)^2 + 5y^2 + 10y - 4z^2 - 15 = 0$

3. Complete the square for $y$ terms

To complete the square for $y$, we need to add and subtract $(10/(2*5))^2 * 5 = 5$ inside the equation:

$10(x - 3)^2 + 5(y^2 + 2y) - 4z^2 - 15 = 0$

$10(x - 3)^2 + 5(y^2 + 2y + 1 - 1) - 4z^2 - 15 = 0$

$10(x - 3)^2 + 5((y + 1)^2 - 1) - 4z^2 - 15 = 0$

$10(x - 3)^2 + 5(y + 1)^2 - 5 - 4z^2 - 15 = 0$

$10(x - 3)^2 + 5(y + 1)^2 - 4z^2 - 20 = 0$

4. Rearrange the equation

Move the constant to the right side:

$10(x - 3)^2 + 5(y + 1)^2 - 4z^2 = 20$

5. Divide by the constant to standardize

Divide both sides by 20 to get 1 on the right side:

$\frac{10(x - 3)^2}{20} + \frac{5(y + 1)^2}{20} - \frac{4z^2}{20} = 1$

$\frac{(x - 3)^2}{2} + \frac{(y + 1)^2}{4} - \frac{z^2}{5} = 1$

6. Identify the quadric surface

The equation is now in the standard form of a quadric surface. Since we have two positive terms and one negative term on the left side of the equation, it represents a hyperboloid of one sheet.