Understanding Eccentricity of Conics: A Fundamental Property

The eccentricity ($e$) is a crucial parameter that defines the shape of a conic section. It quantifies how much a conic section deviates from a perfect circle (for ellipses) or a straight line (for parabolas, where $e=1$). For an ellipse, it measures how "flattened" or "circular" it is, with $0 \le e < 1$. For a hyperbola, it measures its "openness" or "spread," with $e > 1$. This problem requires us to calculate the eccentricities of both an ellipse and a hyperbola, and then use a given relationship between them to find the value of $\theta$.

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Key Concepts and Formulas for Eccentricity

Before diving into the problem, let's review the standard forms and eccentricity formulas for ellipses and hyperbolas centered at the origin.

1. Standard Form of an Ellipse: The general equation of an ellipse centered at the origin is $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$.

   • To find the eccentricity, we need to identify the semi-major axis ($a$) and semi-minor axis ($b$).
   • Identification of $a^2$ and $b^2$:
     • If $A^2 > B^2$, the major axis is along the x-axis. In this case, $a^2 = A^2$ and $b^2 = B^2$.
     • If $B^2 > A^2$, the major axis is along the y-axis. In this case, $a^2 = B^2$ and $b^2 = A^2$.
   • Eccentricity Formula for Ellipse: $e = \sqrt{1 - \frac{b^2}{a^2}}$.
   • Tip: Always remember that $a$ represents the semi-major axis (the larger of the two denominators in the standard form) and $b$ represents the semi-minor axis (the smaller of the two).

2. Standard Form of a Hyperbola: The general equation of a hyperbola centered at the origin can be $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$ or $\frac{y^2}{A^2} - \frac{x^2}{B^2} = 1$.

   • For $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$: The transverse axis is along the x-axis. Here, $a^2 = A^2$ (the denominator of the positive term) and $b^2 = B^2$ (the denominator of the negative term).
   • For $\frac{y^2}{A^2} - \frac{x^2}{B^2} = 1$: The transverse axis is along the y-axis. Here, $a^2 = A^2$ (the denominator of the positive term) and $b^2 = B^2$ (the denominator of the negative term).
   • Eccentricity Formula for Hyperbola: $e = \sqrt{1 + \frac{b^2}{a^2}}$.
   • Tip: For a hyperbola, $a^2$ is always associated with the transverse axis (the term with the positive sign), and $b^2$ is associated with the conjugate axis (the term with the negative sign). Unlike an ellipse, $a$ is not necessarily larger than $b$.

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Step 1: Determine the Eccentricity of the Ellipse ($e_1$)

The given equation of the ellipse is: $x^2 \operatorname{cosec}^2 \theta + y^2 = 5$

Why this step? To apply the eccentricity formula, we must first transform the given equation into its standard form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$. This allows us to correctly identify the squares of the semi-axes, $a^2$ and $b^2$.

1. Convert to Standard Form: To make the right-hand side equal to $1$, we divide the entire equation by $5$: $\frac{x^2 \operatorname{cosec}^2 \theta}{5} + \frac{y^2}{5} = 1$ Now, rewrite the coefficient of $x^2$ in the denominator to match the standard form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$: $\frac{x^2}{\frac{5}{\operatorname{cosec}^2 \theta}} + \frac{y^2}{5} = 1$ Using the trigonometric identity $\frac{1}{\operatorname{cosec}^2 \theta} = \sin^2 \theta$, we simplify the denominator of $x^2$: $\frac{x^2}{5 \sin^2 \theta} + \frac{y^2}{5} = 1$

2. Identify $A^2$ and $B^2$ and Determine Major Axis: From the standard form, we have $A^2 = 5 \sin^2 \theta$ and $B^2 = 5$. Why this comparison? For an ellipse, correctly identifying the major axis (which corresponds to $a^2$) is crucial for using the eccentricity formula. We need to compare $A^2$ and $B^2$. Given the condition $0 < \theta < \pi/2$, we know that $0 < \sin \theta < 1$. Therefore, $0 < \sin^2 \theta < 1$. This implies that $5 \sin^2 \theta < 5$. So, $A^2 < B^2$. This means the denominator under $y^2$ is larger, which indicates that the major axis of the ellipse is along the y-axis.

3. Assign $a_e^2$ and $b_e^2$ and Calculate $e_1$: Since the major axis is along the y-axis, the semi-major axis squared is $a_e^2 = B^2 = 5$. The semi-minor axis squared is $b_e^2 = A^2 = 5 \sin^2 \theta$. The eccentricity of the ellipse, $e_1$, is given by the formula: $e_1 = \sqrt{1 - \frac{b_e^2}{a_e^2}}$ Substitute the values of $a_e^2$ and $b_e^2$: $e_1 = \sqrt{1 - \frac{5 \sin^2 \theta}{5}}$ $e_1 = \sqrt{1 - \sin^2 \theta}$ Using the Pythagorean trigonometric identity $\cos^2 \theta + \sin^2 \theta = 1$, which implies $1 - \sin^2 \theta = \cos^2 \theta$: $e_1 = \sqrt{\cos^2 \theta}$ Since the given domain is $0 < \theta < \pi/2$, $\cos \theta$ is positive in this interval. Therefore, $\sqrt{\cos^2 \theta} = \cos \theta$. $e_1 = \cos \theta$ Common Mistake: Forgetting to consider the sign of $\cos \theta$ when taking the square root. Always use $|\cos \theta|$ and then apply the given domain for $\theta$ to resolve the absolute value.

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Step 2: Determine the Eccentricity of the Hyperbola ($e_2$)

The given equation of the hyperbola is: $x^2 - y^2 \operatorname{cosec}^2 \theta = 5$

Why this step? Similar to the ellipse, we must convert the hyperbola's equation into its standard form to correctly identify $a^2$ (semi-transverse axis squared) and $b^2$ (semi-conjugate axis squared) for the eccentricity calculation.

1. Convert to Standard Form: To make the right-hand side equal to $1$, we divide the entire equation by $5$: $\frac{x^2}{5} - \frac{y^2 \operatorname{cosec}^2 \theta}{5} = 1$ Now, rewrite the coefficient of $y^2$ in the denominator to match the standard form $\frac{x^2}{a_h^2} - \frac{y^2}{b_h^2} = 1$: $\frac{x^2}{5} - \frac{y^2}{\frac{5}{\operatorname{cosec}^2 \theta}} = 1$ Using the identity $\frac{1}{\operatorname{cosec}^2 \theta} = \sin^2 \theta$, we simplify the denominator of $y^2$: $\frac{x^2}{5} - \frac{y^2}{5 \sin^2 \theta} = 1$

2. Identify $a_h^2$ and $b_h^2$: This equation is now in the standard form $\frac{x^2}{a_h^2} - \frac{y^2}{b_h^2} = 1$. This form indicates that the transverse axis is along the x-axis. Here, $a_h^2$ is the denominator of the positive $x^2$ term, so $a_h^2 = 5$. And $b_h^2$ is the denominator of the negative $y^2$ term, so $b_h^2 = 5 \sin^2 \theta$.

3. Calculate $e_2$: The eccentricity of the hyperbola, $e_2$, is given by the formula: $e_2 = \sqrt{1 + \frac{b_h^2}{a_h^2}}$ Substitute the values of $a_h^2$ and $b_h^2$: $e_2 = \sqrt{1 + \frac{5 \sin^2 \theta}{5}}$ $e_2 = \sqrt{1 + \sin^2 \theta}$

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Step 3: Use the Given Relationship to Solve for $\theta$}

The problem states that the eccentricity of the hyperbola ($e_2$) is $\sqrt{7}$ times the eccentricity of the ellipse ($e_1$): $e_2 = \sqrt{7} \cdot e_1$

Why this step? This is the core condition provided in the problem statement that links the two conic sections. By substituting the expressions for $e_1$ and $e_2$ that we derived, we can form an equation solely in terms of $\theta$ and solve for it.

1. Substitute and Form the Equation: Substitute $e_1 = \cos \theta$ (from Step 1) and $e_2 = \sqrt{1 + \sin^2 \theta}$ (from Step 2) into the given relationship: $\sqrt{1 + \sin^2 \theta} = \sqrt{7} \cdot \cos \theta$

2. Solve for $\theta$: To eliminate the square root, square both sides of the equation: $(\sqrt{1 + \sin^2 \theta})^2 = (\sqrt{7} \cdot \cos \theta)^2$ $1 + \sin^2 \theta = 7 \cos^2 \theta$ Why square both sides? Squaring removes the radical, simplifying the equation into a trigonometric identity that is easier to solve.

   Now, convert the equation into a single trigonometric function. We can use the identity $\sin^2 \theta = 1 - \cos^2 \theta$ to express everything in terms of $\cos \theta$: $1 + (1 - \cos^2 \theta) = 7 \cos^2 \theta$ $2 - \cos^2 \theta = 7 \cos^2 \theta$ Move all $\cos^2 \theta$ terms to one side: $2 = 7 \cos^2 \theta + \cos^2 \theta$ $2 = 8 \cos^2 \theta$ Divide by $8$ to solve for $\cos^2 \theta$: $\cos^2 \theta = \frac{2}{8}$ $\cos^2 \theta = \frac{1}{4}$ Take the square root of both sides: $\cos \theta = \pm \sqrt{\frac{1}{4}}$ $\cos \theta = \pm \frac{1}{2}$

   Given the domain $0 < \theta < \pi/2$, $\cos \theta$ must be positive. Therefore, we take the positive value: $\cos \theta = \frac{1}{2}$

   For $0 < \theta < \pi/2$, the angle whose cosine is $\frac{1}{2}$ is $\frac{\pi}{3}$. $\theta = \frac{\pi}{3}$

3. Check the Options: The calculated value $\theta = \frac{\pi}{3}$ matches option (C).

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Summary and Key Takeaway

This problem effectively tests your understanding of:

1. Standard forms of ellipse and hyperbola: The ability to convert general equations into standard forms is fundamental.
2. Identification of $a^2$ and $b^2$: Knowing how to correctly identify the semi-axes (major/minor for ellipse, transverse/conjugate for hyperbola) from the standard form. Remember that for an ellipse, $a^2$ is always the larger denominator, while for a hyperbola, $a^2$ is always under the positive term.
3. Eccentricity formulas: Applying the correct formulas $e = \sqrt{1 - b^2/a^2}$ for ellipse and $e = \sqrt{1 + b^2/a^2}$ for hyperbola.
4. Trigonometric identities and domain constraints: Using $\sin^2