This problem is a classic example that beautifully integrates concepts from coordinate geometry (specifically, ellipses and their tangents) with differential calculus (for optimization). Our ultimate goal is to determine a specific angle $\theta$ in the first quadrant that results in the minimum possible sum of the x and y-intercepts made by a tangent line to the given ellipse.

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1. Understanding the Ellipse and the Parametric Point

The given equation of the ellipse is ${{{x^2}} \over {27}} + {{{y^2}} \over 1} = 1$.

• Why this step? To effectively work with an ellipse, it's crucial to identify its standard parameters ($a$ and $b$) by comparing it with the general standard form of an ellipse centered at the origin: ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$.

By comparing the given equation with the standard form, we can identify:

• $a^2 = 27 \implies a = \sqrt{27} = 3\sqrt{3}$
• $b^2 = 1 \implies b = 1$

The point of tangency is given in parametric form as $(3\sqrt 3 \cos \theta ,\sin \theta )$.

• Why this check? We confirm that this point indeed lies on the ellipse and matches its parametric representation. The general parametric coordinates for an ellipse are $(a \cos \theta, b \sin \theta)$. Substituting our values for $a$ and $b$: $(3\sqrt 3 \cos \theta, 1 \sin \theta)$, which perfectly matches the given point.

The condition $\theta \in \left( {0,{\pi \over 2}} \right)$ is critical. It implies that the point of tangency is in the first quadrant, which means $\sin\theta$, $\cos\theta$, $\tan\theta$, and their reciprocals ($\csc\theta$, $\sec\theta$, $\cot\theta$) will all be positive. This will be important when we perform divisions involving these trigonometric functions later, ensuring they are non-zero and positive.

2. Equation of the Tangent Line to an Ellipse

The most efficient way to find the equation of a tangent to an ellipse when the point of tangency is given in parametric form is to use the dedicated parametric tangent formula.

• Key Concept/Formula: The general equation of a tangent to an ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ at a parametric point $(a \cos \theta, b \sin \theta)$ is given by: ${{x \cos \theta} \over a} + {{y \sin \theta} \over b} = 1$

• Why this formula? This formula is derived by substituting the parametric coordinates $(x_1 = a \cos\theta, y_1 = b \sin\theta)$ into the general tangent equation ${{x{x_1}} \over {{a^2}}} + {{y{y_1}} \over {{b^2}}} = 1$ and simplifying. It streamlines the process significantly compared to using derivatives to find the slope.

Now, we substitute the specific values of $a = 3\sqrt{3}$ and $b = 1$ into this formula: ${{x \cos \theta} \over {3\sqrt 3 }} + {{y \sin \theta} \over 1} = 1$ This is the equation of the tangent line to our given ellipse at the specified parametric point.

3. Calculating Intercepts on the Coordinate Axes

To find the sum of intercepts, we first need to determine the x-intercept and y-intercept of the tangent line.

• Why this step? The problem asks us to minimize the sum of intercepts. Therefore, finding these intercepts is a direct requirement.

• x-intercept: This is the point where the line crosses the x-axis.

  • How to find it? At the x-intercept, the y-coordinate is $0$. So, we set $y=0$ in the tangent equation: ${{x \cos \theta} \over {3\sqrt 3 }} + (0)\sin \theta = 1$ ${{x \cos \theta} \over {3\sqrt 3 }} = 1$
  • Why can we divide? Since $\theta \in \left( {0,{\pi \over 2}} \right)$, $\cos \theta$ is positive and thus non-zero. This allows us to safely divide by $\cos \theta$: $x = {{3\sqrt 3 } \over {\cos \theta }} = 3\sqrt 3 \sec \theta$ So, the x-intercept is $3\sqrt 3 \sec \theta$.

• y-intercept: This is the point where the line crosses the y-axis.

  • How to find it? At the y-intercept, the x-coordinate is $0$. So, we set $x=0$ in the tangent equation: {(0) \cos \theta} \over {3\sqrt 3 }} + y\sin \theta = 1 $y\sin \theta = 1$
  • Why can we divide? Since $\theta \in \left( {0,{\pi \over 2}} \right)$, $\sin \theta$ is positive and thus non-zero. This allows us to safely divide by $\sin \theta$: $y = {1 \over {\sin \theta }} = \csc \theta$ So, the y-intercept is $\csc \theta$.

4. Forming the Function to be Minimized

Let $S(\theta)$ represent the sum of the intercepts on the axes.

• Why this step? We need to define the quantity that we are asked to minimize as a function of our variable, $\theta$. $S(\theta) = (\text{x-intercept}) + (\text{y-intercept})$ $S(\theta) = 3\sqrt 3 \sec \theta + \csc \theta$ Our task is now to find the value of $\theta \in \left( {0,{\pi \over 2}} \right)$ for which this function $S(\theta)$ attains its minimum value.

5. Optimization using Differential Calculus

To find the minimum value of $S(\theta)$, we employ the standard procedure from differential calculus:

• Key Concept: For a function $f(x)$, local minima or maxima occur at critical points where the first derivative $f'(x) = 0$ or is undefined. The second derivative test ($f''(x) > 0$ for minimum, $f''(x) < 0$ for maximum) helps distinguish between them.

Step 5.1: Calculate the First Derivative, $S'(\theta)$

• Why this step? The first derivative tells us about the rate of change of $S(\theta)$. Setting it to zero will help us locate points where the tangent to the curve of $S(\theta)$ is horizontal, which are potential minima or maxima.

Recall the derivatives of the trigonometric functions:

• ${d \over {d\theta}}(\sec \theta) = \sec \theta \tan \theta$
• ${d \over {d\theta}}(\csc \theta) = -\csc \theta \cot \theta$

Now, differentiate $S(\theta)$ with respect to $\theta$: $S'(\theta) = {d \over {d\theta}}(3\sqrt 3 \sec \theta + \csc \theta)$ $S'(\theta) = 3\sqrt 3 (\sec \theta \tan \theta) - (\csc \theta \cot \theta)$

Step 5.2: Set the First Derivative to Zero to Find Critical Points

• Why this step? Setting the first derivative to zero helps us find the critical points where the slope of $S(\theta)$ is zero. These are the candidates for local minima or maxima. $3\sqrt 3 \sec \theta \tan \theta - \csc \theta \cot \theta = 0$ Rearrange the equation: $3\sqrt 3 \sec \theta \tan \theta = \csc \theta \cot \theta$

Step 5.3: Solve the Trigonometric Equation for $\theta$

• Why this strategy? To simplify and solve complex trigonometric equations, it's often easiest to express all trigonometric functions in terms of their fundamental components, $\sin \theta$ and $\cos \theta$.

Convert all terms to $\sin \theta$ and $\cos \theta$: $3\sqrt 3 \left( {1 \over {\cos \theta }} \right) \left( {{\sin \theta } \over {\cos \theta }} \right) = \left( {1 \over {\sin \theta }} \right) \left( {{\cos \theta } \over {\sin \theta }} \right)$ Simplify the expressions on both sides: $3\sqrt 3 {{\sin \theta } \over {{{\cos }^2}\theta }} = {{\cos \theta } \over {{{\sin }^2}\theta }}$ Now, cross-multiply to eliminate denominators: $3\sqrt 3 {\sin^3}\theta = {\cos^3}\theta$

• Why can we divide by $\cos^3\theta$? Since $\theta \in \left( {0,{\pi \over 2}} \right)$, $\cos \theta$ is positive and thus non-zero. Therefore, $\cos^3\theta \neq 0$, and we can safely divide both sides by ${\cos^3}\theta$: $3\sqrt 3 {{\sin^3}\theta \over {{\cos^3}\theta }} = 1$ $3\sqrt 3 {\tan^3}\theta = 1$ Solve for ${\tan^3}\theta$: ${\tan^3}\theta = {1 \over {3\sqrt 3 }}$ To find $\tan\theta$, take the cube root of both sides. It's helpful to recognize that $3\sqrt 3 = \sqrt{9 \times 3} = \sqrt{27}$, or $(3)^{3/2}$. $\tan \theta = {\left( {{1 \over {{(3)^{3/2}}}}} \right)^{1/3}} = {1 \over {{(3)^{(3/2) \times (1/3)}}}} = {1 \over {{(3)^{1/2}}}} = {1 \over {\sqrt 3 }}$

Now, we need to find the angle $\theta$ such that $\tan \theta = {1 \over {\sqrt 3 }}$. For $\theta \in \left( {0,{\pi \over 2}} \right)$, there is a unique value of $\theta$ that satisfies this condition: $\theta = {{\pi \over 6}}$

Step 5.4: Verify Minimum using the Second Derivative Test (Optional but Recommended)

• Why this step? While setting $S'(\theta)=0$ gives critical points, it doesn't distinguish between local minima, local maxima, or inflection points. The second derivative test provides this confirmation. If $S''(\theta) > 0$ at the critical point, it's a local minimum.

Let's find the second derivative, $S''(\theta)$: $S'(\theta) = 3\sqrt 3 \sec \theta \tan \theta - \csc \theta \cot \theta$ Using the product rule and chain rule for differentiation: ${d \over {d\theta}}(\sec \theta \tan \theta) = (\sec \theta \tan \theta)\tan \theta + \sec \theta (\sec^2 \theta) = \sec \theta \tan^2 \theta + \sec^3 \theta$ ${d \over {d\theta}}(-\csc \theta \cot \theta) = -[(-\csc \theta \cot \theta)\cot \theta + \csc \theta (-\csc^2 \theta)] = \csc \theta \cot^2 \theta + \csc^3 \theta$ So, $S''(\theta) = 3\sqrt 3 (\sec \theta \tan^2 \theta + \sec^3 \theta) + (\csc \theta \cot^2 \theta + \csc^3 \theta)$ For $\theta \in \left( {0,{\pi \over 2}} \right)$, all trigonometric functions ($\sec\theta, \tan\theta, \csc\theta, \cot\theta$) are positive. Consequently, all terms in $S''(\theta)$ are positive. Therefore, $S''(\theta) > 0$ for all $\theta \in \left( {0,{\pi \over 2}} \right)$. This confirms that the critical point $\theta = \pi/6$ indeed corresponds to a local minimum for $S(\theta)$.

6. Conclusion

The value of $\theta$ that minimizes the sum of intercepts on the axes made by the tangent is ${{\pi \over 6}}$.

The final answer is $\boxed{{{\pi \over 6}}}$.

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Summary of Key Concepts and Tips for Success:

• Parametric Tangent Equation: For problems involving tangents to ellipses at parametric points, remember the formula $${{x \cos \theta}