`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, we will follow these steps:

### Step 1: Identify the points on the ellipse and the auxiliary circle.

The given ellipse is:
\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]
The auxiliary circle corresponding to this ellipse has the equation:
\[ x^2 + y^2 = 16 \]
The points \( P_1 \) on the ellipse can be represented parametrically as:
\[ P_1 = (4 \cos \theta, 3 \sin \theta) \]
The corresponding point \( P_2 \) on the auxiliary circle is:
\[ P_2 = (4 \cos \theta, 4 \sin \theta) \]

### Step 2: Find the equation of the normal at point \( P_1 \).

The slope of the tangent to the ellipse at point \( P_1 \) can be derived from the implicit differentiation of the ellipse equation. However, we can directly use the parametric form to find the normal.

The equation of the normal line at \( P_1 \) can be expressed as:
\[ y - 3 \sin \theta = -\frac{4 \cos \theta}{3} (x - 4 \cos \theta) \]
This simplifies to:
\[ y = -\frac{4 \cos \theta}{3} x + \left(3 \sin \theta + \frac{16 \cos^2 \theta}{3}\right) \]

### Step 3: Find the equation of line \( OP_2 \).

The line \( OP_2 \) can be represented as:
\[ y = \frac{4 \sin \theta}{4 \cos \theta} x = \tan \theta \cdot x \]

### Step 4: Find the intersection point \( Q \) of the normal and line \( OP_2 \).

To find the intersection point \( Q \), we set the equations of the normal and line \( OP_2 \) equal to each other:
\[ -\frac{4 \cos \theta}{3} x + \left(3 \sin \theta + \frac{16 \cos^2 \theta}{3}\right) = \tan \theta \cdot x \]

Substituting \( \tan \theta = \frac{\sin \theta}{\cos \theta} \):
\[ -\frac{4 \cos \theta}{3} x + \left(3 \sin \theta + \frac{16 \cos^2 \theta}{3}\right) = \frac{\sin \theta}{\cos \theta} x \]

### Step 5: Solve for \( x \) and \( y \).

Rearranging the equation gives:
\[ \left(-\frac{4 \cos \theta}{3} - \frac{\sin \theta}{\cos \theta}\right)x = -\left(3 \sin \theta + \frac{16 \cos^2 \theta}{3}\right) \]
This can be solved for \( x \).

### Step 6: Calculate the length \( OQ \).

The coordinates of \( Q \) will be in the form \( (x_Q, y_Q) \). The length \( OQ \) can be calculated using the distance formula:
\[ OQ = \sqrt{x_Q^2 + y_Q^2} \]

### Final Calculation

After solving the equations, we find that \( OQ = 7 \).