Find the area of the region enclosed by the ellipse \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. \]
Step 1: The area of an ellipse with the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is given by the formula: \[ A = \pi a b. \]
Step 2: Here, \( a \) and \( b \) represent the lengths of the semi-major and semi-minor axes of the ellipse, respectively.
Step 3: Therefore, the area enclosed by the ellipse is simply: \[ A = \pi a b. \] Thus, the area of the region enclosed by the ellipse is \( \pi a b \).
Let \( ABC \) be a triangle formed by the lines \( 7x - 6y + 3 = 0 \), \( x + 2y - 31 = 0 \), and \( 9x - 2y - 19 = 0 \).
Let the point \( (h, k) \) be the image of the centroid of \( \triangle ABC \) in the line \( 3x + 6y - 53 = 0 \). Then \( h^2 + k^2 + hk \) is equal to:
Let \( \overrightarrow{a} = i + 2j + k \) and \( \overrightarrow{b} = 2i + 7j + 3k \).
Let \[ L_1 : \overrightarrow{r} = (-i + 2j + k) + \lambda \overrightarrow{a}, \quad \lambda \in \mathbb{R} \] and \[ L_2 : \overrightarrow{r} = (j + k) + \mu \overrightarrow{b}, \quad \mu \in \mathbb{R} \] be two lines. If the line \( L_3 \) passes through the point of intersection of \( L_1 \) and \( L_2 \), and is parallel to \( \overrightarrow{a} + \overrightarrow{b} \), then \( L_3 \) passes through the point:
Find the shortest distance between the lines \[ \mathbf{r}_1 = (i + 2j + k) + \lambda(i - j + k) \quad \text{and} \quad \mathbf{r}_2 = (2i - j - k) + \mu(2i + j + 2k). \]
Minimize Z = 5x + 3y \text{ subject to the constraints} \[ 4x + y \geq 80, \quad x + 5y \geq 115, \quad 3x + 2y \leq 150, \quad x \geq 0, \quad y \geq 0. \]