Question

If the line \(3x+4y+k = 0\) touches the ellipse \(9x^2+16y^2 = 144\), then the value of k is.

• A. \(\pm 3\sqrt{2}\)
• B. \(\pm 4\sqrt{2}\)
• C. \(\mp 8\sqrt{2}\)
• D. \(\mp 12\sqrt{2}\)

Hint:

Write everything in cos 2 theta and solve the resulting cubic.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Let \(c = \cos 2\theta\). Then \(\cos 4\theta = 2c^2 - 1\). The equation \(\sec4\theta - \sec2\theta = 2\) becomes \(\frac{1}{2c^2 - 1} - \frac{1}{c} = 2\).

Step 2: Form the cubic:
\[ c - (2c^2 - 1) = 2c(2c^2 - 1) \Rightarrow 4c^3 + 2c^2 - 3c - 1 = 0 \]
Check \(c = -1\): \(-4 + 2 + 3 - 1 = 0\). So \((c + 1)(4c^2 - 2c - 1) = 0\).
\[ c = -1\quad\text{or}\quad c = \frac{1 \pm \sqrt{5}}{4} \]

Step 3: Solve for theta:
\(c = -1\): \(2\theta = (2n+1)\pi\), so \(\theta = n\pi + \frac{\pi}{2}\).
\(c = \frac{1 + \sqrt5}{4} = \cos 36^\circ\): \(\theta = n\pi \pm \frac{\pi}{10}\).
\(c = \frac{1 - \sqrt5}{4} = \cos 108^\circ\): \(\theta = n\pi \pm \frac{3\pi}{10}\).
The last two families together give all odd multiples of \(\frac{\pi}{10}\), which is \(\theta = \frac{n\pi}{5} + \frac{\pi}{10}\).

Step 4: Match with the options:
Option (C) lists \(n\pi + \frac{\pi}{2}\) and \(\frac{n\pi}{5} + \frac{\pi}{10}\), which are exactly our two families.

Final Answer:
The solution is \(\theta = n\pi + \frac{\pi}{2},\ \frac{n\pi}{5} + \frac{\pi}{10}\), option (C). \[ \boxed{n\pi + \frac{\pi}{2},\ \frac{n\pi}{5} + \frac{\pi}{10}} \]