`P_(1) and P_(2)` are corresponding points on the ellipse `(x^(2))/(16)+(y^(2))/(9)=1` and its auxiliary circle respectively. If the normal at `P_(1)` to the ellipse meets `OP_(2)` in Q (where O is the origin), then the length of OQ is equal to

Options

A. 3 units
B. 9 units
C. 4 units
D. 7 units

Answer

To solve the problem step by step, we will follow the outlined process to find the length of OQ.

### Step 1: Identify the ellipse and its auxiliary circle
The given ellipse is represented by the equation:
\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]
This ellipse has a semi-major axis \(a = 4\) and a semi-minor axis \(b = 3\). The auxiliary circle corresponding to this ellipse has a radius equal to the semi-major axis, which is \(4\). Its equation is:
\[ x^2 + y^2 = 16 \]

### Step 2: Parametrize the points on the ellipse and the circle
Let \(P_1\) be a point on the ellipse. We can express the coordinates of \(P_1\) in terms of a parameter \(\theta\) as follows:
\[ P_1 = (4 \cos \theta, 3 \sin \theta) \]
The corresponding point \(P_2\) on the auxiliary circle can be expressed as:
\[ P_2 = (4 \cos \theta, 4 \sin \theta) \]

### Step 3: Write the equation of the normal at point \(P_1\)
The equation of the normal at point \(P_1\) on the ellipse is given by:
\[ \frac{a^2}{x} \sec \theta - \frac{b^2}{y} \csc \theta = a^2 - b^2 \]
Substituting \(a = 4\) and \(b = 3\), we have:
\[ 4x \sec \theta - 3y \csc \theta = 16 - 9 \]
This simplifies to:
\[ 4x \sec \theta - 3y \csc \theta = 7 \]

### Step 4: Write the equation of line \(OP_2\)
The line \(OP_2\) can be expressed in slope-intercept form. The slope of line \(OP_2\) is given by:
\[ \text{slope} = \frac{y}{x} = \tan \theta \]
Thus, the equation of line \(OP_2\) is:
\[ y = \tan \theta \cdot x \]

### Step 5: Solve the equations simultaneously
Substituting \(y = \tan \theta \cdot x\) into the normal equation:
\[ 4x \sec \theta - 3(\tan \theta \cdot x) \csc \theta = 7 \]
This can be simplified as:
\[ 4x \sec \theta - 3x \frac{\sin \theta}{\cos \theta} \frac{1}{\sin \theta} = 7 \]
\[ 4x \sec \theta - 3x \frac{1}{\cos \theta} = 7 \]
Factoring out \(x\):
\[ x(4 \sec \theta - 3 \sec \theta) = 7 \]
\[ x = \frac{7 \cos \theta}{\cos \theta} = 7 \cos \theta \]

### Step 6: Find the corresponding \(y\) coordinate
Now substituting \(x = 7 \cos \theta\) back to find \(y\):
\[ y = \tan \theta \cdot (7 \cos \theta) = 7 \sin \theta \]
Thus, the coordinates of point \(Q\) are:
\[ Q = (7 \cos \theta, 7 \sin \theta) \]

### Step 7: Calculate the length \(OQ\)
The length \(OQ\) can be calculated using the distance formula:
\[ OQ = \sqrt{(7 \cos \theta - 0)^2 + (7 \sin \theta - 0)^2} \]
\[ OQ = \sqrt{49 \cos^2 \theta + 49 \sin^2 \theta} \]
Using the identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ OQ = \sqrt{49} = 7 \]

### Final Answer
Thus, the length of \(OQ\) is:
\[ \boxed{7} \]