Understanding the Problem and Key Concepts

We are tasked with finding the value of $5m^2$ given an ellipse $E$ and a line $mx - y = 4$ that is tangent to it. The ellipse's properties (center, focus, vertex) are provided, which will allow us to determine its equation. The core of this problem lies in accurately finding the ellipse's equation and then applying the condition for tangency between a line and an ellipse.

Let's first outline the essential concepts and formulas we'll be using:

1. Standard Equation of an Ellipse:

   • If the major axis is parallel to the x-axis (horizontal), with center $(h, k)$, the equation is: $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ Here, $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis.
   • If the major axis is parallel to the y-axis (vertical), with center $(h, k)$, the equation is: $\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$ In both cases, $a > b$.

2. Geometric Properties of an Ellipse (for a horizontal major axis):

   • Center: $(h, k)$
   • Vertices: $(h \pm a, k)$
   • Foci: $(h \pm ae, k)$, where $e$ is the eccentricity.
   • The distance from the center to a vertex is $a$.
   • The distance from the center to a focus is $ae$.

3. Relationship between $a$, $b$, and $e$: For any ellipse, these parameters are related by the equation: $b^2 = a^2(1-e^2)$

4. Condition for Tangency of a Line to an Ellipse:

   • A line $Y = m'X + c'$ is tangent to the standard ellipse $\frac{X^2}{A^2} + \frac{Y^2}{B^2} = 1$ (which is centered at the origin $(0,0)$) if and only if: $c'^2 = A^2 m'^2 + B^2$
   • This formula is a powerful shortcut, derived from setting the discriminant of the quadratic equation (formed by substituting the line into the ellipse) to zero.

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Step 1: Determine the Equation of the Ellipse $E$

Our first objective is to find the specific values for the center $(h,k)$, the semi-major axis $a$, and the semi-minor axis $b$ (or $b^2$) to construct the ellipse's equation.

1.1 Identify the Center and Orientation of the Major Axis

• Given Information:

  • Center $C = (3, -4)$
  • One Focus $S = (4, -4)$
  • One Vertex $A = (5, -4)$

• Explanation: We observe that all three given points (center, focus, and vertex) share the same y-coordinate, which is $-4$. This means they all lie on the horizontal line $y = -4$. Since the center, foci, and vertices lie on the major axis of an ellipse, this implies that the major axis of ellipse $E$ is parallel to the x-axis.

• Result: The center of the ellipse is $(h, k) = (3, -4)$. Because the major axis is horizontal, the standard form of the ellipse's equation will be $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$.

1.2 Calculate the Semi-Major Axis Length ($a$)

• Key Concept: For an ellipse with a horizontal major axis, the distance from the center $(h, k)$ to a vertex $(h \pm a, k)$ is $a$.
• Explanation: We can directly calculate $a$ by finding the distance between the given center $C(3, -4)$ and the vertex $A(5, -4)$.
• Calculation: $a = \text{Distance}(C, A) = |x_A - x_C| = |5 - 3| = 2$
• Result: The semi-major axis length is $a = 2$. Therefore, $a^2 = 4$.

1.3 Calculate the Eccentricity ($e$)

• Key Concept: For an ellipse with a horizontal major axis, the distance from the center $(h, k)$ to a focus $(h \pm ae, k)$ is $ae$.
• Explanation: We can calculate $ae$ by finding the distance between the given center $C(3, -4)$ and the focus $S(4, -4)$. Since we already know the value of $a$, this will allow us to determine the eccentricity $e$.
• Calculation: $ae = \text{Distance}(C, S) = |x_S - x_C| = |4 - 3| = 1$ Now, substitute the value $a = 2$ that we found earlier: $(2)e = 1 \implies e = \frac{1}{2}$
• Result: The eccentricity of the ellipse is $e = \frac{1}{2}$.

1.4 Calculate the Square of the Semi-Minor Axis Length ($b^2$)

• Key Concept: The fundamental relationship between $a$, $b$, and $e$ for any ellipse is $b^2 = a^2(1-e^2)$.
• Explanation: We have already determined $a=2$ and $e=\frac{1}{2}$. We can now use this relationship to find $b^2$.
• Calculation: $b^2 = (2)^2 \left(1 - \left(\frac{1}{2}\right)^2\right)$ $b^2 = 4 \left(1 - \frac{1}{4}\right)$ $b^2 = 4 \left(\frac{3}{4}\right)$ $b^2 = 3$
• Result: The square of the semi-minor axis length is $b^2 = 3$.

1.5 Write the Equation of Ellipse $E$

• Key Concept: Using the standard equation for a horizontal ellipse centered at $(h,k)$.
• Explanation: We have all the necessary parameters:
  • Center $(h,k) = (3,-4)$
  • $a^2 = 4$
  • $b^2 = 3$ Now, we substitute these into the standard form.
• Equation: $E: \frac{(x-3)^2}{4} + \frac{(y-(-4))^2}{3} = 1$ $E: \frac{(x-3)^2}{4} + \frac{(y+4)^2}{3} = 1$ This is the equation of the ellipse $E$.

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Step 2: Apply the Tangency Condition

We are given the tangent line $mx - y = 4$ (with $m>0$) and need to find $5m^2$. To use the standard tangency condition $c'^2 = A^2 m'^2 + B^2$, we must first transform both the ellipse and the line to a coordinate system where the ellipse is centered at the origin.

2.1 Perform a Coordinate Transformation for the Ellipse

• Why this step is necessary: The tangency formula $c'^2 = A^2 m'^2 + B^2$ is specifically for an ellipse centered at the origin, i.e., of the form $\frac{X^2}{A^2} + \frac{Y^2}{B^2} = 1$. Our ellipse is centered at $(3,-4)$. To apply the formula, we perform a coordinate shift.
• Transformation Definition: Let's define new coordinates $(X, Y)$ such that the origin of this new system coincides with the center of our ellipse $(3, -4)$. $X = x - 3$ $Y = y - (-4) \implies Y = y + 4$
• Ellipse in new coordinates: Substitute these transformations into the equation of ellipse $E$: $\frac{(X)^2}{4} + \frac{(Y)^2}{3} = 1$ Now, this ellipse is centered at $(0,0)$ in the $(X,Y)$ system. By comparing this to the standard form $\frac{X^2}{A^2} + \frac{Y^2}{B^2} = 1$, we identify: $A^2 = 4 \quad \text{and} \quad B^2 = 3$

2.2 Transform the Tangent Line Equation

• Why this step is necessary: For consistency, the tangent line's equation must also be expressed in the new $(X,Y)$ coordinate system. Otherwise, the slope and y-intercept would not correspond to the shifted ellipse.
• Original Tangent Line: The given line is $mx - y = 4$. We can rewrite it in slope-intercept form $y = mx - 4$.
• Substitution using the transformation: We know $x = X + 3$ and $y = Y - 4$. Substitute these into the line equation: $(Y - 4) = m(X + 3) - 4$
• Simplification: $Y - 4 = mX + 3m - 4$ Adding 4 to both sides: $Y = mX + 3m$
• Identify $m'$ and $c'$: Now, compare this transformed line $Y = mX + 3m$ with the general form $Y = m'X + c'$. $m' = m \quad \text{and} \quad c' = 3m$ (Note: Here, $m'$ happens to be the same as the original $m$ because the coordinate transformation is a translation, which does not affect the slope. However, the intercept $c'$ is significantly changed).

2.3 Apply the Tangency Condition

• Key Concept: For the line $Y = m'X + c'$ to be tangent to the ellipse $\frac{X^2}{A^2} + \frac{Y^2}{B^2} = 1$, the condition is $c'^2 = A^2 m'^2 + B^2$.
• Explanation: We now have all the necessary components: $A^2=4$, $B^2=3$, $m'=m$, and $c'=3m$. We can substitute these values into the tangency condition.
• Calculation: $(3m)^2 = (4)(m)^2 + (3)$ $9m^2 = 4m^2 + 3$
• Solve for $5m^2$: $9m^2 - 4m^2 = 3$ $5m^2 = 3$

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Step 3: Final Answer

The problem asks for the value of $5m^2$. From Step 2, we have directly calculated this value.

• Result: $5m^2 = 3$

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Tips and Common Mistakes to Avoid

1. Orientation Check: Always start by checking the orientation of the major axis (horizontal or vertical) from the given points. This dictates which term ($a^2$ or $b^2$) goes under $(x-h)^2$ and $(y-k)^2$. A common error is assuming $a^2$ is always under $x^2$.
2. Consistent Coordinate Transformation: When using the tangency condition for a shifted ellipse, it is absolutely critical to transform both the ellipse equation and the tangent line equation into the new coordinate system. A frequent mistake is transforming only the ellipse and using the original $m$ and $c$ from the line, which will lead to an incorrect answer.
3. Understanding $a$, $b$, and $ae$: Remember their definitions as distances from the center to vertices and foci, respectively. This is key to quickly finding the ellipse's parameters.
4. Discriminant Method (Alternative): While not used here for efficiency, you could substitute $y = mx-4$ directly into the original ellipse equation $\frac{(x-3)^2}{4} + \frac{(y+4)^2}{3} = 1$. This would result in a quadratic equation in $x$. For tangency, the discriminant of this quadratic must be equal to zero. This method is mathematically sound but often involves much more complex algebraic manipulation, especially for shifted conics. The coordinate transformation method is generally more streamlined.
5. Sign Errors: Be careful with signs, especially when dealing with the center coordinates $(h,k)$ and the transformation $Y = y+4$.

Summary and Key Takeaway

This problem effectively tests your ability to:

1. Extract geometric properties of an ellipse from given points (center, focus, vertex).
2. Formulate the standard equation of the ellipse based on these properties.
3. Apply coordinate transformations to simplify the tangency problem for shifted conics.
4. Utilize the tangency condition for a line to an ellipse efficiently.

The coordinate transformation method is a powerful technique for simplifying problems involving conics not centered at the origin, making calculations much cleaner than direct substitution and discriminant evaluation.