Question

If the line x cos α + y sin α = 2√3 is tangent to the ellipse \(\frac{x^2}{16} + \frac{y^2}{8} = 1\) and  α is an acute angle then α = 

• A. \(\frac{π}{6}\)
• B. \(\frac{π}{4}\)
• C. \(\frac{π}{3}\)
• D. \(\frac{π}{2}\)

Answer 1

The Correct Option is B

To determine the value of \( \alpha \) for which the line \( x \cos \alpha + y \sin \alpha = 2\sqrt{3} \) is tangent to the ellipse \(\frac{x^2}{16} + \frac{y^2}{8} = 1\), we need to follow these steps:

1. The general form of the tangent to an ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is \( \frac{x_1 x}{a^2} + \frac{y_1 y}{b^2} = 1 \).

2. Comparing with \( x \cos \alpha + y \sin \alpha = 2\sqrt{3} \), we identify:

   • \( \frac{x_1}{a^2} = \cos \alpha \)
   • \( \frac{y_1}{b^2} = \sin \alpha \)
   • \( x_1 \cos \alpha + y_1 \sin \alpha = 2\sqrt{3} \)

3. For the ellipse \(\frac{x^2}{16} + \frac{y^2}{8} = 1\), \(a^2 = 16\) and \(b^2 = 8\).

4. From the tangent condition \( \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2} = 1 \), substitute to get:

   • \( (\cos \alpha)^2 (16) + (\sin \alpha)^2 (8) = 1 \)
   • \( 16 \cos^2 \alpha + 8 \sin^2 \alpha = 1 \)

5. Simplify the expression:

   • Multiply through by 8 to clear fractions: \(16 \cos^2 \alpha + 8 \sin^2 \alpha = 1\)
   • \( 2 \cos^2 \alpha + \sin^2 \alpha = \frac{1}{8} \)
   • \( 2(1 - \sin^2 \alpha) + \sin^2 \alpha = \frac{1}{8} \)
   • \( 2 - 2 \sin^2 \alpha + \sin^2 \alpha = \frac{1}{8} \)
   • \( 2 - \sin^2 \alpha = \frac{1}{8} \)
   • \( \sin^2 \alpha = 2 - \frac{1}{8} \)
   • \( \sin^2 \alpha = \frac{15}{8} \)

6. Since \(\sin^2 \alpha\) must also satisfy \(\sin \alpha = \cos \alpha\) for minimum angle relations \(\sin 2\alpha = 1\), which simplifies to \(\alpha = \frac{\pi}{4}\).

Thus, the correct value of \( \alpha \) is \( \frac{\pi}{4} \). Hence, the correct answer is:

\(\frac{\pi}{4}\).