If the normal to the ellipse `3x^(2)+4y^(2)=12` at a point P on it is parallel to the line , `2x+y=4` and the tangent to the ellipse at P passes through Q (4,4) then |PQ| is equal to

Options

A. `(5 sqrt(5))/(2)`
B. `( sqrt(61))/(2)`
C. `( sqrt(221))/(2)`
D. `( sqrt(157))/(2)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the equation of the ellipse
The given ellipse is \(3x^2 + 4y^2 = 12\). We can rewrite it in standard form by dividing through by 12:
\[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]
This indicates that \(a^2 = 4\) and \(b^2 = 3\).

### Step 2: Parametric equations of the ellipse
The parametric equations for the ellipse can be expressed as:
\[ x = 2 \cos \theta, \quad y = \sqrt{3} \sin \theta \]
where \(P\) is the point on the ellipse corresponding to the parameter \(\theta\).

### Step 3: Find the equation of the normal at point \(P\)
The equation of the normal to the ellipse at point \(P(x_1, y_1)\) is given by:
\[ \frac{a^2 (x - x_1)}{x_1} + \frac{b^2 (y - y_1)}{y_1} = a^2 - b^2 \]
Substituting \(x_1 = 2 \cos \theta\) and \(y_1 = \sqrt{3} \sin \theta\):
\[ \frac{4(x - 2 \cos \theta)}{2 \cos \theta} + \frac{3(y - \sqrt{3} \sin \theta)}{\sqrt{3} \sin \theta} = 1 \]
This simplifies to:
\[ 2x \sin \theta - \sqrt{3} \cos \theta \cdot y = \sin \theta \cos \theta \]

### Step 4: Find the slope of the normal
The slope of the normal line is given by:
\[ m_1 = \frac{2 \sin \theta}{\sqrt{3} \cos \theta} \]
The line \(2x + y = 4\) can be rewritten in slope-intercept form as:
\[ y = -2x + 4 \]
Thus, the slope of this line is \(m_2 = -2\).

### Step 5: Set the slopes equal
Since the normal is parallel to the line \(2x + y = 4\), we set the slopes equal:
\[ \frac{2 \sin \theta}{\sqrt{3} \cos \theta} = -2 \]
Cross-multiplying gives:
\[ 2 \sin \theta = -2\sqrt{3} \cos \theta \]
This simplifies to:
\[ \sin \theta = -\sqrt{3} \cos \theta \]
Dividing both sides by \(\cos \theta\) (assuming \(\cos \theta \neq 0\)):
\[ \tan \theta = -\sqrt{3} \]
This implies:
\[ \theta = \frac{5\pi}{3} \quad \text{or} \quad \theta = \frac{2\pi}{3} \]

### Step 6: Find the coordinates of point \(P\)
Using \(\theta = \frac{5\pi}{3}\):
\[ x = 2 \cos\left(\frac{5\pi}{3}\right) = 2 \cdot \frac{1}{2} = 1 \]
\[ y = \sqrt{3} \sin\left(\frac{5\pi}{3}\right) = \sqrt{3} \cdot \left(-\frac{\sqrt{3}}{2}\right) = -\frac{3}{2} \]
So, point \(P\) is \((1, -\frac{3}{2})\).

### Step 7: Find the equation of the tangent at point \(P\)
The equation of the tangent line at \(P\) is given by:
\[ \frac{x_1 x}{a^2} + \frac{y_1 y}{b^2} = 1 \]
Substituting \(x_1 = 1\) and \(y_1 = -\frac{3}{2}\):
\[ \frac{1 \cdot x}{4} + \frac{-\frac{3}{2} \cdot y}{3} = 1 \]
This simplifies to:
\[ \frac{x}{4} - \frac{y}{2} = 1 \]
or
\[ x - 2y = 4 \]

### Step 8: Find the intersection with point \(Q(4, 4)\)
To find the distance \(|PQ|\), we can calculate the distance between points \(P(1, -\frac{3}{2})\) and \(Q(4, 4)\):
\[ |PQ| = \sqrt{(4 - 1)^2 + \left(4 - \left(-\frac{3}{2}\right)\right)^2} \]
Calculating:
\[ |PQ| = \sqrt{3^2 + \left(4 + \frac{3}{2}\right)^2} = \sqrt{9 + \left(\frac{8 + 3}{2}\right)^2} = \sqrt{9 + \left(\frac{11}{2}\right)^2} \]
\[ = \sqrt{9 + \frac{121}{4}} = \sqrt{\frac{36}{4} + \frac{121}{4}} = \sqrt{\frac{157}{4}} = \frac{\sqrt{157}}{2} \]

### Final Answer
Thus, the distance \(|PQ|\) is:
\[ |PQ| = \frac{\sqrt{157}}{2} \]