This problem is a classic application of the properties of ellipses, specifically involving the equations of its tangent and normal lines. Our strategy will be to first standardize the ellipse equation and represent the point P using parametric coordinates. Then, we will use the given conditions (normal parallel to a line, tangent passing through a point) to find the coordinates of P. Finally, we will calculate the distance between P and Q.

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1. Understanding the Ellipse Equation and Parametric Coordinates

The first step is to convert the given ellipse equation into its standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. This allows us to identify the semi-major axis ($a$) and semi-minor axis ($b$), which are crucial for defining parametric coordinates and for the equations of tangents and normals.

• Given Ellipse Equation: $3x^2 + 4y^2 = 12$
• Standard Form: To obtain the standard form, we divide the entire equation by 12: $\frac{3x^2}{12} + \frac{4y^2}{12} = \frac{12}{12}$ $\frac{x^2}{4} + \frac{y^2}{3} = 1$
• Identifying Parameters: By comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we can identify: $a^2 = 4 \implies a = 2$ $b^2 = 3 \implies b = \sqrt{3}$
• Parametric Coordinates of P: For any point P on the ellipse, its coordinates can be expressed in parametric form as $(a \cos \theta, b \sin \theta)$. This representation is particularly useful because it inherently satisfies the ellipse equation and simplifies the derivation of tangent and normal equations. So, point P is $(2 \cos \theta, \sqrt{3} \sin \theta)$.

Tip: Using parametric coordinates often simplifies calculations for tangents and normals compared to using general Cartesian coordinates $(x_1, y_1)$.

2. Utilizing the Normal Condition

We are given that the normal to the ellipse at point P is parallel to the line $2x + y = 4$. Parallel lines have equal slopes. We will use this property to find a relationship involving $\theta$.

• Slope of the Given Line: The line is $2x + y = 4$, which can be rewritten as $y = -2x + 4$. The slope of this line, $m_{line}$, is $-2$.
• Equation of the Normal: The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at a point $P(a \cos \theta, b \sin \theta)$ is given by: $\frac{a^2 x}{a \cos \theta} - \frac{b^2 y}{b \sin \theta} = a^2 - b^2$ Substituting $a^2=4$ and $b^2=3$: $\frac{4x}{2 \cos \theta} - \frac{3y}{\sqrt{3} \sin \theta} = 4 - 3$ $\frac{2x}{\cos \theta} - \frac{\sqrt{3}y}{\sin \theta} = 1$
• Slope of the Normal ($m_{normal}$): To find the slope, we rearrange the normal equation into $y = mx + c$ form: $-\frac{\sqrt{3}y}{\sin \theta} = 1 - \frac{2x}{\cos \theta}$ $y = -\frac{\sin \theta}{\sqrt{3}} \left(1 - \frac{2x}{\cos \theta}\right)$ $y = \frac{2 \sin \theta}{\sqrt{3} \cos \theta} x - \frac{\sin \theta}{\sqrt{3}}$ So, the slope of the normal is $m_{normal} = \frac{2 \sin \theta}{\sqrt{3} \cos \theta} = \frac{2}{\sqrt{3}} \tan \theta$.
• Equating Slopes: Since the normal is parallel to $2x+y=4$, their slopes must be equal: $m_{normal} = m_{line}$ $\frac{2}{\sqrt{3}} \tan \theta = -2$ $\tan \theta = -\sqrt{3}$

Explanation: We found $\tan \theta = -\sqrt{3}$. This means $\theta$ could be in the second quadrant (e.g., $120^\circ$ or $2\pi/3$) or the fourth quadrant (e.g., $300^\circ$ or $5\pi/3$). We need another condition to pinpoint the exact value of $\theta$.

Common Mistake: A common mistake is to confuse the formula for the slope of the normal with the slope of the tangent. For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at $(x_1, y_1)$, the slope of the tangent is $-\frac{b^2 x_1}{a^2 y_1}$, and the slope of the normal is $\frac{a^2 y_1}{b^2 x_1}$. Using parametric coordinates, these become $-\frac{b}{a} \cot \theta$ and $\frac{a}{b} \tan \theta$ respectively.

3. Utilizing the Tangent Condition

We are given that the tangent to the ellipse at P passes through the point Q(4,4). We will use the equation of the tangent at P and substitute Q's coordinates to get another equation involving $\theta$.

• Equation of the Tangent: The equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at a point $P(a \cos \theta, b \sin \theta)$ is given by: $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$ Substituting $a=2$ and $b=\sqrt{3}$: $\frac{x \cos \theta}{2} + \frac{y \sin \theta}{\sqrt{3}} = 1$
• Tangent Passes Through Q(4,4): Since the tangent passes through Q(4,4), we substitute $x=4$ and $y=4$ into the tangent equation: $\frac{4 \cos \theta}{2} + \frac{4 \sin \theta}{\sqrt{3}} = 1$ $2 \cos \theta + \frac{4 \sin \theta}{\sqrt{3}} = 1$ $2\sqrt{3} \cos \theta + 4 \sin \theta = \sqrt{3} \quad \text{(Equation 1)}$

Explanation: This equation provides a second relationship between $\sin \theta$ and $\cos \theta$. We can now combine this with the result from the normal condition ($\tan \theta = -\sqrt{3}$) to find the specific values of $\sin \theta$ and $\cos \theta$.

4. Determining the Coordinates of Point P

We have two pieces of information:

1. $\tan \theta = -\sqrt{3}$
2. $2\sqrt{3} \cos \theta + 4 \sin \theta = \sqrt{3}$

From $\tan \theta = -\sqrt{3}$, we know that $\sin \theta$ and $\cos \theta$ must have opposite signs. Also, since $\tan \theta = \frac{\sin \theta}{\cos \theta}$, we can write $\sin \theta = -\sqrt{3} \cos \theta$.

Substitute $\sin \theta = -\sqrt{3} \cos \theta$ into Equation 1: $2\sqrt{3} \cos \theta + 4 (-\sqrt{3} \cos \theta) = \sqrt{3}$ $2\sqrt{3} \cos \theta - 4\sqrt{3} \cos \theta = \sqrt{3}$ $-2\sqrt{3} \cos \theta = \sqrt{3}$ $\cos \theta = -\frac{\sqrt{3}}{2\sqrt{3}}$ $\cos \theta = -\frac{1}{2}$

Now, find $\sin \theta$ using $\sin \theta = -\sqrt{3} \cos \theta$: $\sin \theta = -\sqrt{3} \left(-\frac{1}{2}\right)$ $\sin \theta = \frac{\sqrt{3}}{2}$

Verification: We have $\cos \theta = -1/2$ and $\sin \theta = \sqrt{3}/2$. $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\sqrt{3}/2}{-1/2} = -\sqrt{3}$. This is consistent with our finding from the normal condition. Also, $\cos^2 \theta + \sin^2 \theta = (-1/2)^2 + (\sqrt{3}/2)^2 = 1/4 + 3/4 = 1$, which confirms these are valid trigonometric values. This particular combination of $\cos \theta = -1/2$ and $\sin \theta = \sqrt{3}/2$ corresponds to $\theta = 120^\circ$ or $2\pi/3$, which is in the second quadrant.

• Coordinates of P: Now substitute these values back into the parametric coordinates of P: $P(2 \cos \theta, \sqrt{3} \sin \theta)$ $P\left(2 \left(-\frac{1}{2}\right), \sqrt{3} \left(\frac{\sqrt{3}}{2}\right)\right)$ $P\left(-1, \frac{3}{2}\right)$

Explanation: By combining the two conditions, we uniquely determined the trigonometric values of $\theta$, which in turn gave us the precise coordinates of point P.

5. Calculating the Distance PQ

Finally, we need to find the distance between point P and point Q.

• Point P is $\left(-1, \frac