This problem is a comprehensive test of your understanding of various conic sections: circles, parabolas, and ellipses. It requires you to apply tangent conditions for different curves, solve systems of equations, and then use the properties of an ellipse to find its eccentricity and latus rectum. We will proceed step-by-step, explaining the rationale behind each action.

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1. Understanding the Problem and Key Concepts

Our goal is to find the value of ${l \over {{e^2}}}$ for a specific ellipse. To do this, we'll follow these main stages:

• Stage 1: Find the Common Tangents. We'll identify the given circle and parabola, write down their standard forms, and then use the general tangent equations to find the common tangents.
• Stage 2: Locate the Intersection Point Q. Once we have the equations of the common tangents, we'll find their intersection point, Q.
• Stage 3: Determine Ellipse Parameters. We'll calculate the distance OQ (where O is the origin) and use it, along with the given value of 6, to define the semi-major and semi-minor axes of the ellipse.
• Stage 4: Calculate Ellipse Properties. Finally, we'll calculate the eccentricity ($e$) and the length of the latus rectum ($l$) for this ellipse and compute the desired ratio.

Throughout this process, we will rely on the following fundamental formulas and definitions from coordinate geometry:

• Tangent to a Parabola: For a parabola $y^2 = 4ax$, the equation of a tangent with slope $m$ is $y = mx + \frac{a}{m}$. This formula ensures that the line touches the parabola at exactly one point.
• Tangent to a Circle: For a circle $x^2 + y^2 = r^2$ centered at the origin, the equation of a tangent with slope $m$ is $y = mx \pm r\sqrt{1+m^2}$. This is derived from the condition that the perpendicular distance from the center to the tangent line equals the radius.
• Distance Formula: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
• Ellipse Properties (centered at origin): For an ellipse with semi-major axis $a$ and semi-minor axis $b$ (where $a \ge b$):
  • Eccentricity ($e$): $e^2 = 1 - \frac{b^2}{a^2}$. This value indicates how "oval" the ellipse is.
  • Length of Latus Rectum ($l$): $l = \frac{2b^2}{a}$. This is the length of a chord passing through a focus and perpendicular to the major axis.

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2. Step-by-Step Solution

2.1 Standardizing the Given Curves

The first step is to recognize the given equations and convert them into their standard forms. This allows us to easily identify their key parameters, which are essential for applying the tangent formulas.

1. The Circle: The given equation is $4({x^2} + {y^2}) = 9$. To get it into the standard form $x^2 + y^2 = r^2$, we divide both sides by 4: $x^2 + y^2 = \frac{9}{4}$ This can be rewritten as: $x^2 + y^2 = \left(\frac{3}{2}\right)^2$ Why this step? From this standard form, we can immediately identify that the circle is centered at the origin $(0,0)$ and has a radius $r = \frac{3}{2}$. This radius will be used in the tangent condition for the circle.

2. The Parabola: The given equation is ${y^2} = 4x$. This is already in the standard form $y^2 = 4ax$. Why this step? By comparing $y^2 = 4x$ with $y^2 = 4ax$, we can see that $4a = 4$, which implies $a = 1$. This value of 'a' is critical for writing the tangent equation for the parabola.

2.2 Deriving Tangency Conditions for Each Curve

We are looking for lines that are tangent to both the circle and the parabola. Let's assume the general equation of such a common tangent line is $y = mx + c$, where $m$ is the slope and $c$ is the y-intercept. We will use the specific tangency conditions for each curve to relate $m$ and $c$.

• Tangent to the Parabola $y^2 = 4x$ (with $a=1$): Using the standard tangent formula $y = mx + \frac{a}{m}$, we substitute $a=1$: $y = mx + \frac{1}{m}$ Why this step? This equation represents any line with slope $m$ that is tangent to the parabola $y^2=4x$. By comparing this with our general tangent form $y=mx+c$, we establish a condition for $c$: $c = \frac{1}{m} \quad \text{(Equation 1)}$

• Tangent to the Circle $x^2 + y^2 = \left(\frac{3}{2}\right)^2$ (with $r=\frac{3}{2}$): Using the standard tangent formula $y = mx \pm r\sqrt{1+m^2}$, we substitute $r=\frac{3}{2}$: $y = mx \pm \frac{3}{2}\sqrt{1+m^2}$ Why this step? Similarly, this equation represents any line with slope $m$ that is tangent to the circle. Comparing it with $y=mx+c$, we get the condition for $c$: $c = \pm \frac{3}{2}\sqrt{1+m^2} \quad \text{(Equation 2)}$

2.3 Finding the Slopes of the Common Tangents

Since the line $y=mx+c$ must be a common tangent, the value of $c$ must satisfy both tangency conditions simultaneously. Therefore, we equate the expressions for $c$ from Equation 1 and Equation 2: $\frac{1}{m} = \pm \frac{3}{2}\sqrt{1+m^2}$ Why this step? This is the core idea for finding common tangents: a line satisfying the tangency condition for one curve must also satisfy it for the other. This equation now allows us to solve for the unknown slope $m$.

To solve for $m$, we need to eliminate the square root. We do this by squaring both sides of the equation: $\left(\frac{1}{m}\right)^2 = \left(\pm \frac{3}{2}\sqrt{1+m^2}\right)^2$ $\frac{1}{m^2} = \frac{9}{4}(1+m^2)$ Tip: Squaring both sides of an equation can sometimes introduce extraneous solutions (solutions that satisfy the squared equation but not the original one). It's good practice to verify solutions, especially when dealing with square roots and $\pm$ signs. In this case, since we are solving for $m^2$, any negative value for $m^2$ will be an extraneous solution as $m$ must be real.

Now, we simplify and rearrange the equation: Multiply both sides by $4m^2$ to clear the denominators: $4 = 9m^2(1+m^2)$ $4 = 9m^2 + 9m^4$ Rearrange this into a quadratic equation in terms of $m^2$: $9(m^2)^2 + 9m^2 - 4 = 0$ Let $X = m^2$ to make the quadratic form more explicit: $9X^2 + 9X - 4 = 0$ We solve this quadratic equation for $X$ using the quadratic formula, $X = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$: $X = \frac{-9 \pm \sqrt{9^2 - 4(9)(-4)}}{2(9)}$ $X = \frac{-9 \pm \sqrt{81 + 144}}{18}$ $X = \frac{-9 \pm \sqrt{225}}{18}$ $X = \frac{-9 \pm 15}{18}$ This yields two possible values for $X$: $X_1 = \frac{-9 + 15}{18} = \frac{6}{18} = \frac{1}{3}$ $X_2 = \frac{-9 - 15}{18} = \frac{-24}{18} = -\frac{4}{3}$ Why we discard $X_2$: Since $X = m^2$, and $m$ represents a real slope, $m^2$ must be a non-negative value. Therefore, we must discard $X_2 = -\frac{4}{3}$. Thus, we take the valid solution: $m^2 = \frac{1}{3}$ Taking the square root, we find the possible slopes for the common tangents: $m = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}}$

2.4 Determining the Equations of the Common Tangents

Now that we have the values for $m$, we can find the corresponding $c$ values using Equation 1 ($c = \frac{1}{m}$) and write down the equations of the common tangents. Why this step? We need the full equations of the tangents to find their intersection point.

• Case 1: $m = \frac{1}{\sqrt{3}}$ $c = \frac{1}{1/\sqrt{3}} = \sqrt{3}$ The equation of the first common tangent $L_1$ is: $L_1: y = \frac{1}{\sqrt{3}}x + \sqrt{3}$

• Case 2: $m = -\frac{1}{\sqrt{3}}$ $c = \frac{1}{-1/\sqrt{3}} = -\sqrt{3}$ The equation of the second common tangent $L_2$ is: $L_2: y = -\frac{1}{\sqrt{3}}x - \sqrt{3}$

2.5 Finding the Intersection Point Q and Length OQ

The problem states that the common tangents intersect at point Q. To find the coordinates of Q, we solve the system of linear equations formed by $L_1$ and $L_2$. Why this step? The point Q is crucial for determining one of the semi-axes of the ellipse.

$L_1: y = \frac{1}{\sqrt{3}}x + \sqrt{3}$ $L_2: y = -\frac{1}{\sqrt{3}}x - \sqrt{3}$ Equating the expressions for $y$ from both equations: $\frac{1}{\sqrt{3}}x + \sqrt{3} = -\frac{1}{\sqrt{3}}x - \sqrt{3}$ Gather $x$ terms on one side and constant terms on the other: $\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}x = -\sqrt{3} - \sqrt{3}$ $\frac{2}{\sqrt{3}}x = -2\sqrt{3}$ Multiply both sides by $\frac{\sqrt{3}}{2}$ to solve for $x$: $x = -2\sqrt{3} \times \frac{\sqrt{3}}{2}$ $x = -3$ Now substitute $x=-3$ into either tangent equation (let's use $L_1$): $y = \frac{1}{\sqrt{3}}(-3) + \sqrt{3}$ $y = -\frac{3}{\sqrt{3}} + \sqrt{3}$ $y = -\sqrt{3} + \sqrt{3}$ $y = 0$ So, the intersection point Q is $(-3, 0)$.

Next, we need to find the length of the segment OQ, where O is the origin $(0,0)$. This length will be one of the semi-axes of the ellipse. Why this step? The problem explicitly states that OQ is a semi-axis length.

Using the distance formula between $O(0,0)$ and $Q(-3,0)$: $OQ = \sqrt{(-3 - 0)^2 + (0 - 0)^2}$ $OQ = \sqrt{(-3)^2 + 0^2}$ $OQ = \sqrt{9}$ $OQ = 3$

2.6 Defining the Ellipse and its Semi-Axes

The problem states that an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. Why this step? Correctly identifying the semi-major and semi-minor axes is crucial for accurate calculations of eccentricity and latus rectum.

• We found $OQ = 3$.
• The other given length is 6.

By definition, the