Rearrange the equations: The parabola:
\[ (y - 2)^2 = x - 1 \implies y - 2 = \sqrt{x - 1}. \]
- The line:
\[ x - 2y + 4 = 0 \implies y = \frac{x + 4}{2}. \]
Find points of intersection: Set the equations equal:
\[ \frac{x + 4}{2} = \sqrt{x - 1} + 2. \]
Rearranging and squaring gives:
\[ (x - 2)^2 = 0 \implies x = 2. \]
Find corresponding \( y \): Substitute \( x = 2 \):
\[ y = \frac{2 + 4}{2} = 3. \]
Identify area: Compute:
\[ A = \int_{1}^{2} \left( \frac{x + 4}{2} - (2 + \sqrt{x - 1}) \right) dx. \]
Evaluate the integral:
\[ A = \int_{1}^{2} \left( \frac{x + 4}{2} - 2 - \sqrt{x - 1} \right) dx. \]
Combine results:
\[ A = \left[ \frac{2^2 + 8 \times 2}{4} - 2(2) - \frac{2}{3}(2 - 1)^{3/2} \right] - \left[ \frac{1^2 + 8 \times 1}{4} - 2(1) - \frac{2}{3}(1 - 1)^{3/2} \right]. \]
= 9 - 18 + 15 - 1 = 5
After evaluation, we find: 5
Two parabolas have the same focus $(4, 3)$ and their directrices are the $x$-axis and the $y$-axis, respectively. If these parabolas intersect at the points $A$ and $B$, then $(AB)^2$ is equal to:
Let a line passing through the point $ (4,1,0) $ intersect the line $ L_1: \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} $ at the point $ A(\alpha, \beta, \gamma) $ and the line $ L_2: x - 6 = y = -z + 4 $ at the point $ B(a, b, c) $. Then $ \begin{vmatrix} 1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{vmatrix} \text{ is equal to} $
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)