1. Introduction to Tangent Equations for Conic Sections

The problem asks for the square of the slope of a common tangent line to two curves: an ellipse and a circle, both centered at the origin. To solve this, we will use the standard equations for tangent lines to these conic sections.

For a line $y = mx + c$ to be tangent to a conic section centered at the origin, the value of $c$ (the y-intercept) is determined by the slope $m$ and the parameters of the conic.

• Tangent to an Ellipse: For an ellipse in standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the equation of a tangent line with slope $m$ is given by: $y = mx \pm \sqrt{a^2m^2 + b^2}$
• Tangent to a Circle: For a circle centered at the origin in standard form $x^2 + y^2 = r^2$, the equation of a tangent line with slope $m$ is given by: $y = mx \pm r\sqrt{1+m^2}$

Our strategy will be to:

1. Convert both given curve equations into their standard forms.
2. Apply the respective tangent formulas to express the y-intercept $c$ in terms of $m$ for each curve.
3. Since the line L is a common tangent, its y-intercept must be the same for both curves for a given slope $m$. We will equate these expressions for $c$.
4. Solve the resulting equation for $m^2$.

2. Step 1: Standardizing the Equations of the Curves

Before applying the tangent formulas, we must convert the given equations into their standard forms to identify the parameters $a^2, b^2,$ and $r^2$.

For the Ellipse: Given equation: $4x^2 + 9y^2 = 36$

• Why this step? The general tangent formula for an ellipse requires the equation to be in the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. This step allows us to correctly identify $a^2$ and $b^2$. To achieve the standard form, divide the entire equation by 36: $\frac{4x^2}{36} + \frac{9y^2}{36} = \frac{36}{36}$ $\frac{x^2}{9} + \frac{y^2}{4} = 1$ Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we identify: $a^2 = 9 \quad \text{and} \quad b^2 = 4$

For the Circle: Given equation: $(2x)^2 + (2y)^2 = 31$

• Why this step? The general tangent formula for a circle requires the equation to be in the form $x^2 + y^2 = r^2$. This step allows us to correctly identify the radius squared, $r^2$. First, simplify the squares: $4x^2 + 4y^2 = 31$ Now, divide the entire equation by 4 to get $x^2 + y^2 = r^2$: $x^2 + y^2 = \frac{31}{4}$ Comparing this with the standard form $x^2 + y^2 = r^2$, we identify: $r^2 = \frac{31}{4}$ From this, the radius $r$ is $r = \sqrt{\frac{31}{4}} = \frac{\sqrt{31}}{2}$.

3. Step 2: Formulating Tangent Equations with Slope $m$

Now we substitute the identified parameters into the general tangent formulas. Let the slope of the common tangent line be $m$.

Tangent to the Ellipse: Using $y = mx \pm \sqrt{a^2m^2 + b^2}$ with $a^2=9$ and $b^2=4$: $y = mx \pm \sqrt{9m^2 + 4} \quad \text{...(1)}$ Here, the y-intercept $c_1 = \pm \sqrt{9m^2 + 4}$.

Tangent to the Circle: Using $y = mx \pm r\sqrt{1+m^2}$ with $r = \frac{\sqrt{31}}{2}$: $y = mx \pm \frac{\sqrt{31}}{2}\sqrt{1+m^2} \quad \text{...(2)}$ Here, the y-intercept $c_2 = \pm \frac{\sqrt{31}}{2}\sqrt{1+m^2}$.

4. Step 3: Equating Y-Intercepts for a Common Tangent

• Why this step? A line L is a common tangent to both curves. This means that for a specific slope $m$, the line $y=mx+c$ must be identical for both curves. Therefore, the y-intercepts (the constant terms) from equations (1) and (2) must be equal in magnitude. If a line is tangent to both, it means it has the same slope AND the same y-intercept for both conditions. We equate the expressions for the y-intercepts: $\sqrt{9m^2 + 4} = \frac{\sqrt{31}}{2}\sqrt{1+m^2}$ Note: We only need to equate the positive parts because squaring both sides will inherently account for the $\pm$ signs. If $c_1 = \pm c_2$, then $c_1^2 = c_2^2$.

5. Step 4: Solving for the Square of the Slope ($m^2$)

Now, we solve the equation from Step 3 to find the value of $m^2$.

• Why this step? The problem specifically asks for the square of the slope, $m^2$. This algebraic manipulation allows us to isolate $m^2$. Square both sides of the equation to eliminate the square roots: $(\sqrt{9m^2 + 4})^2 = \left(\frac{\sqrt{31}}{2}\sqrt{1+m^2}\right)^2$ $9m^2 + 4 = \frac{31}{4}(1+m^2)$ To eliminate the fraction, multiply the entire equation by 4: $4(9m^2 + 4) = 4 \times \frac{31}{4}(1+m^2)$ $36m^2 + 16 = 31(1+m^2)$ Distribute 31 on the right side: $36m^2 + 16 = 31 + 31m^2$ Now, rearrange the terms to group $m^2$ terms on one side and constants on the other: $36m^2 - 31m^2 = 31 - 16$ $5m^2 = 15$ Finally, solve for $m^2$: $m^2 = \frac{15}{5}$ $m^2 = 3$

6. Final Answer

The square of the slope of the common tangent line L is $m^2 = 3$.

7. Tips and Common Mistakes

• Standard Form is Paramount: Always convert the conic section equations to their standard forms before applying any formulas. A common mistake is to use incorrect $a^2, b^2,$ or $r^2$ values if the equations are not properly standardized.
• Algebraic Precision: Pay close attention to algebraic manipulations, especially when squaring both sides of an equation or distributing terms. Small errors here can lead to incorrect results.
• Understanding the Common Tangent Condition: The core idea is that the specific line $y=mx+c$ must satisfy tangency conditions for both curves simultaneously. This means both $m$ and $c$ must be consistent. Equating the magnitude of the y-intercepts for