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 find the acute angle \(\alpha\) 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 proceed as follows:

1. Identifying Parameters of the Ellipse:
The ellipse is given by \(\frac{x^2}{16} + \frac{y^2}{8} = 1\), which is in the form \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\). 
\( a^2 = 16 \implies a = 4 \)
\( b^2 = 8 \implies b = 2\sqrt{2} \)

2. Line Equation:
The line is given by:
\( x \cos \alpha + y \sin \alpha = 2\sqrt{3} \)
Rewrite in standard form:
\( x \cos \alpha + y \sin \alpha - 2\sqrt{3} = 0 \)

3. Tangency Condition for the Ellipse:
For a line \( x \cos \alpha + y \sin \alpha = p \) to be tangent to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), 
\( a^2 \cos^2 \alpha + b^2 \sin^2 \alpha = p^2 \)
Here, \( a^2 = 16 \), \( b^2 = 8 \), and \( p = 2\sqrt{3} \), so \( p^2 = (2\sqrt{3})^2 = 12 \). 
\( 16 \cos^2 \alpha + 8 \sin^2 \alpha = 12 \)

4. Simplifying the Equation:
Use \(\sin^2 \alpha = 1 - \cos^2 \alpha\):
\( 16 \cos^2 \alpha + 8 (1 - \cos^2 \alpha) = 12 \)
\( 16 \cos^2 \alpha + 8 - 8 \cos^2 \alpha = 12 \)
\( 8 \cos^2 \alpha + 8 = 12 \)
\( 8 \cos^2 \alpha = 4 \)
\( \cos^2 \alpha = \frac{4}{8} = \frac{1}{2} \)

5. Solving for \(\alpha\):
\( \cos \alpha = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2} \)
Since \(\alpha\) is an acute angle (\(0 < \alpha < \frac{\pi}{2}\)), take the positive root:
\( \cos \alpha = \frac{\sqrt{2}}{2} \implies \alpha = \frac{\pi}{4} \)

Final Answer:
The acute angle \(\alpha\) is \(\frac{\pi}{4}\).