To find the slope of a common tangent to two curves, we generally follow a strategy that involves expressing the general equation of a tangent for one curve (usually the one with a simpler tangent formula) and then applying the tangency condition for the second curve. This approach allows us to determine the unknown parameters of the tangent line, such as its slope.

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1. Understanding the Given Curves and Their Tangency Conditions

We are given two conic sections: an ellipse and a circle. Let's analyze each one and recall the relevant formulas for their tangents.

• Curve 1: The Ellipse ($C_1$) The equation is $\frac{x^2}{16} + \frac{y^2}{9} = 1$ Why this form? This is the standard form of an ellipse centered at the origin, $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. By comparing, we identify its parameters: $a^2 = 16$ $b^2 = 9$ Key Formula for Ellipse Tangency: The equation of a tangent line with slope $m$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by: $y = mx \pm \sqrt{a^2m^2 + b^2}$ Why this formula? This formula is a fundamental result in coordinate geometry. It can be derived using calculus (by finding points where the derivative $\frac{dy}{dx}$ equals $m$) or by setting the discriminant of the quadratic equation formed by intersecting the line $y = mx + c$ with the ellipse to zero (which implies only one point of intersection, characteristic of a tangent). It's incredibly useful for problems where the slope $m$ is the primary unknown.

• Curve 2: The Circle ($C_2$) The equation is $x^2 + y^2 = 12$ Why this form? This is the standard form of a circle centered at the origin, $(x-h)^2 + (y-k)^2 = R^2$. By comparing, we identify its properties: Center $(h, k) = (0, 0)$ (the origin) Radius squared $R^2 = 12$, so the radius $R = \sqrt{12}$. Key Condition for Circle Tangency: A line $Ax + By + C = 0$ is tangent to a circle with center $(h, k)$ and radius $R$ if and only if the perpendicular distance from the center of the circle to the line is equal to the radius. The distance formula for a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$ For tangency, we set $d = R$: $R = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}}$ Why this condition? This is the geometric definition of a tangent to a circle: it touches the circle at exactly one point, and the radius drawn to that point is perpendicular to the tangent. The distance formula mathematically captures this perpendicularity.

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2. Formulating the Equation of a Tangent to the Ellipse ($C_1$)

Let $m$ be the slope of the common tangent line. We will use the standard formula for the tangent to the ellipse $C_1$: $y = mx \pm \sqrt{a^2m^2 + b^2}$ Substitute the specific values for our ellipse, $a^2 = 16$ and $b^2 = 9$: $y = mx \pm \sqrt{16m^2 + 9}$ Explanation: This equation now represents any line with slope $m$ that is tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$. The $\pm$ sign indicates that for a given slope $m$, there are generally two such tangent lines (one on each side of the ellipse). Our next step is to find the specific value(s) of $m$ for which this line is also tangent to the circle.

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3. Applying the Tangency Condition for the Circle ($C_2$)

For the line $y = mx \pm \sqrt{16m^2 + 9}$ to be tangent to the circle $C_2$, the perpendicular distance from the center of $C_2$ to this line must be equal to the radius of $C_2$.

First, we need to rewrite the tangent equation into the general form $Ax + By + C = 0$ to apply the distance formula conveniently. $y = mx \pm \sqrt{16m^2 + 9}$ Rearranging the terms, we get: $mx - y \pm \sqrt{16m^2 + 9} = 0$ Explanation: This form allows us to directly identify the coefficients $A$, $B$, and $C$ for the distance formula.

• $A = m$
• $B = -1$
• $C = \pm \sqrt{16m^2 + 9}$ (Note: The $\pm$ sign here comes from the tangent equation itself, indicating two possible constant terms for the line).

Now, we apply the distance formula for tangency to the circle $C_2$: The center of the circle is $(h, k) = (0, 0)$ and its radius is $R = \sqrt{12}$. $R = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}}$ Substitute the identified values: $\sqrt{12} = \frac{|m(0) - 1(0) \pm \sqrt{16m^2 + 9}|}{\sqrt{m^2 + (-1)^2}}$ Simplifying the expression: $\sqrt{12} = \frac{|\pm \sqrt{16m^2 + 9}|}{\sqrt{m^2 + 1}}$ Explanation of Absolute Value: The term $\sqrt{16m^2 + 9}$ represents a non-negative real number. The absolute value of $\pm X$ is simply $|X|$. Therefore, $|\pm \sqrt{16m^2 + 9}|$ simplifies to $\sqrt{16m^2 + 9}$. So, the equation becomes: $\sqrt{12} = \frac{\sqrt{16m^2 + 9}}{\sqrt{m^2 + 1}}$

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4. Solving the Resulting Equation for $m^2$

To eliminate the square roots and solve for $m^2$, we square both sides of the equation: $\left( \frac{\sqrt{16m^2 + 9}}{\sqrt{m^2 + 1}} \right)^2 = (\sqrt{12})^2$ Why square both sides? Squaring is a standard algebraic technique to remove square roots, allowing us to work with a polynomial equation. This simplifies to: $\frac{16m^2 + 9}{m^2 + 1} = 12$ Now, we perform algebraic manipulation to isolate $m^2$: Multiply both sides by $(m^2 + 1)$ to clear the denominator: $16m^2 + 9 = 12(m^2 + 1)$ Distribute the 12 on the right side: $16m^2 + 9 = 12m^2 + 12$ Gather all terms involving $m^2$ on one side and constant terms on the other side: $16m^2 - 12m^2 = 12 - 9$ $4m^2 = 3$

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5. Calculating the Required Value

The question asks for the value of $12m^2$, not just $m^2$ or $m$. From the previous step, we found the relationship: $4m^2 = 3$ Why this step? It's crucial to always refer back to what the question asks for. Solving for $m^2$ and then multiplying is more efficient than solving for $m$ (which would be $\pm \sqrt{3}/2$) and then squaring and multiplying. To obtain $12m^2$, we can multiply both sides of this equation by 3: $3 \times (4m^2) = 3 \times 3$ $12m^2 = 9$

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Tips for Success and Common Pitfalls:

• Memorize Standard Formulas: Having the tangent equation for ellipses and the tangency condition for circles (distance from center = radius) readily available is crucial for solving such problems quickly and accurately.
• Careful Algebraic Manipulation: Pay close attention to signs, especially when rearranging the line equation into $Ax+By+C=0$. Errors in signs or squaring can lead to incorrect results.
• Understand Absolute Values: Remember that $|\pm K| = |K|$. This simplification step is important and often a source of confusion for students.
• Check the Final Question: Always double-check what the question is specifically asking for. Here, it was $12m^2$, not $m^2$ or $m$. Many students might stop prematurely at $m^2 = 3/4$.
• Geometric Intuition (Optional but helpful): Visualize the curves. An ellipse and a circle generally have four common tangents (two with positive slopes, two with negative slopes). Since $m^2$ is what we solved for, it naturally accounts for both positive and negative slopes (as $(-m)^2 = m^2$).

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Summary and Key Takeaway:

This problem is a classic application of analytical geometry, combining the properties of an ellipse and a circle. The most efficient approach involves:

1. Using the standard tangent equation for the ellipse in terms of its slope $m$.
2. Rewriting this tangent line equation into the general form $Ax+By+C=0$.
3. Imposing the tangency condition for the circle, which states that the perpendicular distance from the circle's center to the tangent line must equal its radius.
4. Solving the resulting algebraic equation for $m^2$.
5. Finally, calculating the specific value requested by the problem ($12m^2$).

The final calculation yields $12m^2 = 9$.

The correct option is (B).