Question:hard

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

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Write everything in cos 2 theta and solve the resulting cubic.
Updated On: Oct 1, 2026
  • \(\pm 3\sqrt{2}\)
  • \(\pm 4\sqrt{2}\)
  • \(\mp 8\sqrt{2}\)
  • \(\mp 12\sqrt{2}\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Test the listed solutions:
Try $\theta = \frac{\pi}{2}$: $\sec 2\pi = 1$ and $\sec\pi = -1$, so the difference is $1 - (-1) = 2$. It works.
Try $\theta = \frac{\pi}{10}$: $\sec72^\circ - \sec36^\circ = 3.236 - 1.236 = 2$. It works.
Try $\theta = \frac{3\pi}{10}$: $\sec216^\circ - \sec108^\circ = -1.236 + 3.236 = 2$. It works.

Step 2: Why the other options fail:
For option (A), $\theta = \frac{\pi}{8}$ gives $\sec\frac{\pi}{2}$, which is undefined. For option (B), $\theta = \frac{\pi}{6}$ gives $\sec120^\circ - \sec60^\circ = -2 - 2 = -4$. For option (D), $\theta = \frac{\pi}{3}$ gives $\sec240^\circ - \sec120^\circ = 0$. So only (C) is right.

Final Answer:
Option (C). \[ \boxed{\text{(C)}} \]
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