1. Conceptual Framework: Understanding the Problem and Key Tools

This problem asks us to find the acute angle between two tangent lines drawn to a given ellipse from an external point. This is a classic analytical geometry problem that combines the properties of ellipses with fundamental concepts of straight lines.

To solve this, we will follow a structured approach involving these key mathematical tools:

• Standard Form of an Ellipse: An ellipse centered at the origin is typically represented by the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. This form is crucial because it directly provides the parameters $a^2$ and $b^2$, which are essential for subsequent formulas.
• Equation of a Tangent in Slope Form: For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the equation of a tangent line with a slope $m$ is given by: $y = mx \pm \sqrt{a^2m^2 + b^2}$ This formula is powerful because it allows us to relate the slope of a tangent to the ellipse's parameters.
• Angle Between Two Lines: If two lines have slopes $m_1$ and $m_2$, the angle $\theta$ between them can be calculated using the formula: $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$ The absolute value sign is critical here; it ensures that we always obtain a non-negative value for $\tan \theta$, which corresponds to the acute angle between the lines (i.e., $\theta \in [0, \pi/2]$).
• Vieta's Formulas: For a quadratic equation $Am^2 + Bm + C = 0$ whose roots are $m_1$ and $m_2$, Vieta's formulas provide a direct way to find the sum and product of the roots without explicitly solving the quadratic:
  • Sum of roots: $m_1 + m_2 = -\frac{B}{A}$
  • Product of roots: $m_1 m_2 = \frac{C}{A}$ These formulas are extremely efficient for our problem, as we only need the sum and product of the slopes for the angle formula.

Our Strategy:

1. Convert the given ellipse equation to its standard form to identify $a^2$ and $b^2$.
2. Substitute the coordinates of the external point into the slope form of the tangent equation to form a quadratic equation in $m$. The roots of this quadratic will be the slopes of the two tangents.
3. Use Vieta's formulas to find the sum ($m_1+m_2$) and product ($m_1m_2$) of these slopes.
4. Apply the angle formula $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$ to compute the required angle.

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2. Step 1: Standardizing the Ellipse Equation

The given equation of the ellipse is: $2 x^{2}+3 y^{2}=5$ Our first goal is to transform this into the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. This allows us to directly identify the values of $a^2$ and $b^2$, which are the fundamental parameters of the ellipse used in tangent equations.

To achieve this, we divide every term in the equation by $5$: $\frac{2x^2}{5} + \frac{3y^2}{5} = \frac{5}{5}$ $\frac{x^2}{5/2} + \frac{y^2}{5/3} = 1$ From this standard form, we can clearly identify: $a^2 = \frac{5}{2} \quad \text{and} \quad b^2 = \frac{5}{3}$ Why this step? The tangent equation in slope form, $y = mx \pm \sqrt{a^2m^2 + b^2}$, explicitly requires $a^2$ and $b^2$. Standardizing the ellipse equation is the most straightforward way to extract these values.

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3. Step 2: Deriving the Quadratic Equation for Tangent Slopes

We know that the two tangents pass through the external point $(1,3)$. We will use the general slope form of the tangent equation and substitute this point's coordinates to find the specific slopes for these tangents.

The equation of a tangent with slope $m$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is: $y = mx \pm \sqrt{a^2m^2 + b^2}$ Now, substitute the coordinates of the external point $(x,y) = (1,3)$ and the values of $a^2 = 5/2$ and $b^2 = 5/3$: $3 = m(1) \pm \sqrt{\left(\frac{5}{2}\right)m^2 + \frac{5}{3}}$ $3 - m = \pm \sqrt{\frac{5}{2}m^2 + \frac{5}{3}}$ Why this step? The external point $(1,3)$ must lie on any tangent drawn from it. By substituting its coordinates, we introduce a crucial constraint that allows us to determine the specific slopes ($m$) of the tangents passing through this point.

To eliminate the square root and obtain a polynomial equation, we square both sides of the equation: $(3 - m)^2 = \left(\pm \sqrt{\frac{5}{2}m^2 + \frac{5}{3}}\right)^2$ $9 - 6m + m^2 = \frac{5}{2}m^2 + \frac{5}{3}$ Why squaring? Squaring both sides is necessary to remove the radical, converting the equation into a polynomial form, specifically a quadratic equation in $m$. We expect a quadratic equation because from any external point, exactly two tangents can be drawn to an ellipse, implying there will be two distinct slopes ($m_1$ and $m_2$). Common Mistake Alert: Always isolate the square root term before squaring both sides. If you square $3 - m = \pm \sqrt{...}$ as $(3-m)^2 = (\pm \sqrt{...})^2$, it's correct. But if you had something like $A + \sqrt{B} = C$ and squared it as $(A+\sqrt{B})^2 = C^2$, you would still have a radical term ($2A\sqrt{B}$) and wouldn't simplify properly.

Now, we rearrange the terms to form a standard quadratic equation $Am^2 + Bm + C = 0$: $m^2 - \frac{5}{2}m^2 - 6m + 9 - \frac{5}{3} = 0$ Combine the $m^2$ terms and the constant terms: $\left(1 - \frac{5}{2}\right)m^2 - 6m + \left(9 - \frac{5}{3}\right) = 0$ $\left(\frac{2-5}{2}\right)m^2 - 6m + \left(\frac{27-5}{3}\right) = 0$ $-\frac{3}{2}m^2 - 6m + \frac{22}{3} = 0$ To simplify and work with integer coefficients, which is generally easier and less prone to calculation errors, we can multiply the entire equation by $-6$ (which is the least common multiple of the denominators $2$ and $3$, and also makes the leading coefficient positive): $(-6) \left(-\frac{3}{2}m^2 - 6m + \frac{22}{3}\right) = (-6)(0)$ $9m^2 + 36m - 44 = 0$ This is our quadratic equation in $m$. Its roots, $m_1$ and $m_2$, are the slopes of the two tangents drawn from $(1,3)$ to the ellipse.

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4. Step 3: Utilizing Vieta's Formulas for Slopes

For the quadratic equation $9m^2 + 36m - 44 = 0$, we identify the coefficients: $A = 9$ $B = 36$ $C = -44$

Let the roots of this equation be $m_1$ and $m_2$, representing the slopes of the two tangents. Using Vieta's formulas, we can find their sum and product:

• Sum of roots: $m_1 + m_2 = -\frac{B}{A} = -\frac{36}{9} = -4$
• Product of roots: $m_1 m_2 = \frac{C}{A} = \frac{-44}{9}$

Why Vieta's Formulas? We only need the sum and product of the slopes for the angle formula in the next step, not the individual values of $m_1$ and $m_2$. Vieta's formulas allow us to obtain these values directly and efficiently from the coefficients of the quadratic equation, saving us the effort of solving the quadratic equation explicitly (which might involve complex square roots if we were to use the quadratic formula).

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5. Step 4: Calculating the Acute Angle Between the Tangents

The formula for the acute angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is: $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$ Why the absolute value? The absolute value ensures that $\tan \theta$ is positive, which by convention, gives us the acute angle ($0 \le \theta \le \pi/2$).

We already have $m_1 + m_2 = -4$ and $m_1 m_2 = -\frac{44}{9}$. To use the angle formula, we first need to find the value of $|m_1 - m_2|$. We can achieve this using the algebraic identity: $(m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1 m_2$ Substitute the known values: $(m_1 - m_2)^2 = (-4)^2 - 4\left(-\frac{44}{9}\right)$ $(m_1 - m_2)^2 = 16 + \frac{176}{9}$ To add these terms, find a common denominator: