Question

For \(0 < \theta < \frac{pi}{2}\), if the eccentricity of the hyperbola \( x^2 - y^2 \csc^2 \theta = 5 \) is \( \sqrt{7} \) times the eccentricity of the ellipse \( x^2 \csc^2 \theta + y^2 = 5 \), then the value of \( \theta \) is:

• A. \(\frac{\pi}{6}\)
• B. \(\frac{5\pi}{12}\)
• C. \(\frac{\pi}{3}\)
• D. \(\frac{\pi}{4}\)

Answer 1

The Correct Option is C

For the hyperbola, the eccentricity is given by:

\[ e_h = \sqrt{1 + \sin^2 \theta} \]

For the ellipse, the eccentricity is given by:

\[ e_e = \sqrt{1 - \sin^2 \theta} \]

We are given that the eccentricity of the hyperbola is \(\sqrt{7}\) times the eccentricity of the ellipse:

\[ e_h = \sqrt{7} \cdot e_e \]

Substituting the expressions for \(e_h\) and \(e_e\):

\[ \sqrt{1 + \sin^2 \theta} = \sqrt{7} \cdot \sqrt{1 - \sin^2 \theta} \]

Squaring both sides:

\[ 1 + \sin^2 \theta = 7(1 - \sin^2 \theta) \]

Expanding:

\[ 1 + \sin^2 \theta = 7 - 7 \sin^2 \theta \]

Simplifying:

\[ 1 + \sin^2 \theta + 7 \sin^2 \theta = 7 \]

\[ 8 \sin^2 \theta = 6 \]

\[ \sin^2 \theta = \frac{3}{4} \]

Thus:

\[ \sin \theta = \frac{\sqrt{3}}{2} \]

Therefore:

\[ \theta = \frac{\pi}{3} \]

Answer 2

For \( 0 < \theta < \frac{\pi}{2} \), given that the eccentricity of the hyperbola \( x^2 - y^2 \csc^2 \theta = 5 \) is \( \sqrt{7} \) times the eccentricity of the ellipse \( x^2 \csc^2 \theta + y^2 = 5 \), find \( \theta \).

Concept Used:

For a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), eccentricity \( e_h = \sqrt{1 + \frac{b^2}{a^2}} \).
For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) (with \( a > b \)), eccentricity \( e_e = \sqrt{1 - \frac{b^2}{a^2}} \).

Step-by-Step Solution:

Step 1: Write both equations in standard form.

Hyperbola: \( x^2 - y^2 \csc^2 \theta = 5 \)
Divide by 5: \( \frac{x^2}{5} - \frac{y^2}{5 \sin^2 \theta} = 1 \).
So \( a_h^2 = 5 \), \( b_h^2 = 5 \sin^2 \theta \).

Ellipse: \( x^2 \csc^2 \theta + y^2 = 5 \)
Divide by 5: \( \frac{x^2}{5 \sin^2 \theta} + \frac{y^2}{5} = 1 \).
Here, denominator of \( x^2 \) is \( 5 \sin^2 \theta \), of \( y^2 \) is 5.
Since \( 0 < \theta < \frac{\pi}{2} \), \( 0 < \sin \theta < 1 \), so \( 5 \sin^2 \theta < 5 \).
Thus \( a_e^2 = 5 \) (semi-major axis along y-axis), \( b_e^2 = 5 \sin^2 \theta \).

Step 2: Find eccentricities.

Hyperbola: \( e_h = \sqrt{1 + \frac{b_h^2}{a_h^2}} = \sqrt{1 + \frac{5 \sin^2 \theta}{5}} = \sqrt{1 + \sin^2 \theta} \).

Ellipse: \( e_e = \sqrt{1 - \frac{b_e^2}{a_e^2}} = \sqrt{1 - \frac{5 \sin^2 \theta}{5}} = \sqrt{1 - \sin^2 \theta} = \sqrt{\cos^2 \theta} = \cos \theta \) (since \( \cos \theta > 0 \) in given domain).

Step 3: Use given condition.

Given: \( e_h = \sqrt{7} \, e_e \)
So \( \sqrt{1 + \sin^2 \theta} = \sqrt{7} \cos \theta \).

Step 4: Solve for \( \sin^2 \theta \).

Square both sides: \( 1 + \sin^2 \theta = 7 \cos^2 \theta \).
But \( \cos^2 \theta = 1 - \sin^2 \theta \), so:
\( 1 + \sin^2 \theta = 7(1 - \sin^2 \theta) \)
\( 1 + \sin^2 \theta = 7 - 7 \sin^2 \theta \)
\( 8 \sin^2 \theta = 6 \)
\( \sin^2 \theta = \frac{3}{4} \).

Step 5: Find \( \theta \).

\( \sin \theta = \frac{\sqrt{3}}{2} \) (since \( \sin \theta > 0 \))
So \( \theta = \frac{\pi}{3} \).

Therefore, the value of \( \theta \) is \( \mathbf{\frac{\pi}{3}} \).