1. Understanding the Ellipse and its Key Properties

An ellipse is a locus of points for which the sum of the distances from two fixed points (foci) is constant. Its properties are best understood from its standard form equations.

The general equation of a conic section is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$. For an ellipse, we typically have $B=0$ and $A, C$ have the same sign (and $A \neq C$ for a non-circular ellipse).

The standard forms of an ellipse centered at $(h,k)$ are:

• Horizontal Ellipse: Major axis parallel to the x-axis. $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$
• Vertical Ellipse: Major axis parallel to the y-axis. $\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2} = 1$

In both cases, $a$ represents the length of the semi-major axis and $b$ represents the length of the semi-minor axis. A fundamental condition for an ellipse is that $a > b$.

From these semi-axis lengths, we derive other important properties:

• Length of Major Axis ($l$): $2a$
• Length of Latus Rectum: $\frac{2b^2}{a}$

Our primary goal is to transform the given general equation of the ellipse into one of these standard forms. This transformation will allow us to identify $a$ and $b$, which are essential for calculating the required quantities.

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2. Converting the General Equation to Standard Form

We are given the equation of an ellipse: $x^{2}+4 y^{2}+2 x+8 y-\lambda=0$

The most effective strategy to convert a general quadratic equation of a conic section into its standard form is by completing the square for the $x$ terms and $y$ terms.

Step 2.1: Grouping and Rearranging Terms First, group the terms involving $x$ and $y$ separately, and move the constant term to the right side of the equation. $(x^2+2x) + (4y^2+8y) = \lambda$

• Why this step? Grouping terms makes it easier to apply the completing the square method independently for the $x$ and $y$ parts of the equation. Moving the constant prepares the equation for the standard form, which has a constant on the right side.

Step 2.2: Completing the Square for $x$ terms To complete the square for $x^2+2x$, we need to add $(\frac{\text{coefficient of } x}{2})^2$. The coefficient of $x$ is 2, so we add $(2/2)^2 = 1^2 = 1$. $(x^2+2x+1)$ This expression is now a perfect square trinomial, $(x+1)^2$.

• Why this step? A perfect square trinomial can be factored into the form $(x-h)^2$, which is required for the standard form of an ellipse. To maintain the equality of the equation, any value added to one side must also be added to the other side. So, we add 1 to both sides of our main equation.

Step 2.3: Completing the Square for $y$ terms For the $y$ terms, we have $4y^2+8y$. Before completing the square, it is crucial to factor out the coefficient of $y^2$ (which is 4) from the $y$ terms. $4(y^2+2y)$ Now, complete the square for the expression inside the parenthesis, $y^2+2y$. Similar to the $x$ terms, add $(2/2)^2 = 1^2 = 1$ inside the parenthesis. $4(y^2+2y+1)$ This expression inside the parenthesis is now $(y+1)^2$. So, we have $4(y+1)^2$.

• Why factor first? The completing the square formula $X^2+BX+(\frac{B}{2})^2 = (X+\frac{B}{2})^2$ only applies when the coefficient of $X^2$ is 1. Factoring out the 4 ensures this condition.
• Careful Accounting: When we added 1 inside the parenthesis, it was effectively multiplied by the 4 that was factored out. Therefore, we have actually added $4 \times 1 = 4$ to the left side of the equation. We must also add this value to the right side to keep the equation balanced.

Step 2.4: Substitute Back and Simplify Substitute the completed square forms back into the equation from Step 2.1, remembering to add the balancing constants to the right side: $(x+1)^2 + 4(y+1)^2 = \lambda + 1 \text{ (from x-terms)} + 4 \text{ (from y-terms)}$ $(x+1)^2 + 4(y+1)^2 = \lambda + 5$

• Why this step? This brings us closer to the standard form by grouping the squared terms on one side and all constants on the other.

Step 2.5: Normalize to Standard Form The standard form of an ellipse requires the right-hand side of the equation to be 1. To achieve this, divide the entire equation by the constant term on the right side, which is $(\lambda+5)$. $\frac{(x+1)^2}{\lambda+5} + \frac{4(y+1)^2}{\lambda+5} = 1$ To match the standard form $\frac{(y-k)^2}{B^2}$, we need to rewrite the second term. The coefficient 4 in the numerator can be moved to the denominator of the denominator: $\frac{(x+1)^2}{\lambda+5} + \frac{(y+1)^2}{\left(\frac{\lambda+5}{4}\right)} = 1$

• Why this step? This is the final step to achieve the standard form, which clearly presents the denominators as $a^2$ and $b^2$.
• Common Mistake: Forgetting to divide the coefficient of the squared term when normalizing. Remember that $\frac{C \cdot Y^2}{K} = \frac{Y^2}{K/C}$.

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3. Identifying Semi-Axes ($a$ and $b$) and Orientation

From the standard form we just derived: $\frac{(x+1)^2}{\lambda+5} + \frac{(y+1)^2}{\left(\frac{\lambda+5}{4}\right)} = 1$ For an ellipse, the denominators must be positive, so $\lambda+5 > 0$. We need to identify $a^2$ and $b^2$. Recall that $a$ is the semi-major axis, and $b$ is the semi-minor axis, meaning $a > b$. Therefore, $a^2$ is always the larger of the two denominators, and $b^2$ is the smaller one.

Let's compare the two denominators: $D_x = \lambda+5$ and $D_y = \frac{\lambda+5}{4}$. Since $\lambda+5$ must be positive, it's clear that $\lambda+5$ is greater than $\frac{\lambda+5}{4}$.

Therefore:

• $a^2 = \lambda+5$
• $b^2 = \frac{\lambda+5}{4}$

The fact that $a^2$ is under the $(x+1)^2$ term indicates that the major axis of this ellipse is parallel to the x-axis, making it a horizontal ellipse.

From these, we can find the lengths of the semi-axes:

• $a = \sqrt{\lambda+5}$
• $b = \sqrt{\frac{\lambda+5}{4}} = \frac{\sqrt{\lambda+5}}{2}$
• Why this step? Correctly identifying $a^2$ and $b^2$ (and consequently $a$ and $b$) is absolutely critical because all other properties of the ellipse (like latus rectum and major axis length) depend directly on these values. The condition $a>b$ helps determine which denominator is $a^2$.

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4. Using the Latus Rectum to Find $\lambda$

We are given that the length of the latus rectum of the ellipse is 4. The formula for the length of the latus rectum is: $L.R. = \frac{2b^2}{a}$ Now, substitute the expressions for $a$ and $b^2$ that we found in Step 3: $L.R. = \frac{2 \left(\frac{\lambda+5}{4}\right)}{\sqrt{\lambda+5}}$ Simplify the numerator: $L.R. = \frac{\frac{\lambda+5}{2}}{\sqrt{\lambda+5}}$ We can further simplify this expression. Recall that for any positive number $X$, $\frac{X}{\sqrt{X}} = \sqrt{X}$. Applying this, where $X = \lambda+5$: $L.R. = \frac{1}{2} \cdot \frac{\lambda+5}{\sqrt{\lambda+5}} = \frac{\sqrt{\lambda+5}}{2}$

• Why this step? By substituting the derived expressions for $a$ and $b^2$ into the latus rectum formula, we create an equation solely in terms of $\lambda$. Solving this equation will give us the value of $\lambda$. Algebraic simplification makes the equation easier to solve.

Now, set this simplified expression for the latus rectum equal to the given value, 4: $\frac{\sqrt{\lambda+5}}{2} = 4$ To solve for $\lambda$: Multiply both sides by 2: $\sqrt{\lambda+5} = 8$ Square both sides to eliminate the square root: $(\sqrt{\lambda+5})^2 = 8^2$ $\lambda+5 = 64$ Subtract 5 from both sides: $\lambda = 64 - 5$ $\lambda = 59$

• Why this step? This directly uses the given information to solve for the unknown parameter $\lambda$, which is a crucial part of the problem.

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5. Calculating the Length of the Major Axis ($l$)

The length of the major axis is denoted by $l$, and its formula is $l = 2a$. From Step 3, we know that $a = \sqrt{\lambda+5}$. Now, substitute the value of $\lambda=59$ that we just found into the expression for $a$: $a = \sqrt{59+5}$ $a = \sqrt{64}$ $a = 8$ Now, calculate the length of the major axis, $l$: $l = 2a = 2 \times 8$ $l = 16$

• Why this step? This calculates the second required quantity in the problem statement.

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6. Final Calculation: $\lambda+l$

The question asks for the value of $\lambda+l$. We found $\lambda =