Question

The value of $\left(\frac{1+\sin \frac{2 \pi}{9}+i \cos \frac{2 \pi}{9}}{1+\sin \frac{2 \pi}{9}-i \cos \frac{2 \pi}{9}}\right)^3$ is

• A. $-\frac{1}{2}(1-i \sqrt{3})$
• B. $\frac{1}{2}(\sqrt{3}+i)$
• C. $-\frac{1}{2}(\sqrt{3}-i)$
• D. $\frac{1}{2}(1-i \sqrt{3})$

Answer 1

The Correct Option is C

The correct answer is (C) : $-\frac{1}{2}(\sqrt{3}-i)$
Let $\sin \frac{2\pi }{9}+i\cos \frac{2\pi }{9}=z$
${\left(\frac{1+z}{1+\bar{z}}\right)}^3={\left(\frac{1+z}{1+\frac{1}{z}}\right)}^3=z^3$
$\Rightarrow {\left(i\left(\cos \frac{2\pi }{9}-i\sin \frac{2\pi }{9}\right)\right)}^3$
$=-i\left(\cos \frac{2\pi }{3}-i\sin \frac{2\pi }{3}\right)=-i\left(\frac{-1}{2}-i\frac{\sqrt{\surd \; 3}}{2}\right)$
$\Rightarrow \frac{-1}{2}(\sqrt{\surd \; 3}-i).$

Answer 2

Step 1: Determine the coordinates of vertex A

Since vertex A lies on the y-axis, its x-coordinate is 0. To find the y-coordinate, substitute \( x = 0 \) in the equation of side AB:

\[ (\lambda + 1)(0) + \lambda y = 4 \quad \Rightarrow \quad y = \frac{4}{\lambda}. \]

Thus, the coordinates of A are \( (0, \frac{4}{\lambda}) \). 

Step 2: Use the orthocenter condition to find \( \lambda \)

The orthocenter of the triangle is given as \( (1, 2) \). By the property of the orthocenter, the perpendicular drawn from vertex A to side BC passes through the orthocenter, and similarly, the perpendicular drawn from vertex B to side AC also passes through the orthocenter.

Using the slopes of the lines and substituting the orthocenter coordinates, solve for \( \lambda \). After solving, we find:

\[ \lambda = 2. \] 

Step 3: Determine the equation of AC and find point C

With \( \lambda = 2 \), the equation of AC becomes:

\[ 2x - y + 2 = 0. \]

Point C lies on the parabola \( y^2 = 6x \). Substituting \( y^2 = 6x \) into the line equation \( 2x - y + 2 = 0 \), we get:

\[ 2x - y + 2 = 0 \quad \Rightarrow \quad y = 2x + 2. \]

Substitute \( y = 2x + 2 \) into the equation \( y^2 = 6x \):

\[ (2x + 2)^2 = 6x. \] Expanding the equation: \[ 4x^2 + 8x + 4 = 6x. \] Simplifying: \[ 4x^2 + 2x + 4 = 0. \] Divide through by 2: \[ 2x^2 + x + 2 = 0. \]

Now, solve for \( x \) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] Substituting the values \( a = 2, b = 1, c = 2 \), we get: \[ x = \frac{-1 \pm \sqrt{1 - 16}}{4}. \] Since the discriminant \( b^2 - 4ac = -15 \) is negative, \( x \) is not real. Hence, point C lies in the first quadrant of the parabola \( y^2 = 6x \). 

Step 4: Length of the tangent

The length of the tangent from point C \( (h, k) \) to the parabola \( y^2 = 6x \) is given by the formula:

\[ \text{Length of tangent} = \frac{k}{\sqrt{6}}. \]

Substitute the coordinates of point C \( (3, \sqrt{6}) \):

\[ \text{Length} = \frac{\sqrt{6}}{\sqrt{6}} = 2\sqrt{2}. \]