If the normal at any point P of the ellipse `(x^(2))/(16)+(y^(2))/(9) =1` meets the coordinate axes at M and N respectively, then `|PM|: |PN|` equals

Options

A. `4:3`
B. `16:9`
C. `9:16`
D. `3:4`

Explanation

To solve the problem, we need to find the ratio of the distances from the point \( P \) on the ellipse to the points \( M \) and \( N \), where the normal at \( P \) intersects the coordinate axes.

### Step 1: Find the equation of the normal at point \( P \)

The equation of the ellipse is given by:

\[ \frac{x^2}{16} + \frac{y^2}{9} = 1 \]

The semi-major axis \( a = 4 \) and semi-minor axis \( b = 3 \).

The parametric equations for the ellipse can be expressed as:

\[ x = 4 \cos \theta, \quad y = 3 \sin \theta \]

The equation of the normal at point \( P \) (where \( P = (4 \cos \theta, 3 \sin \theta) \)) is given by:

\[ a x \sec \theta - b y \cos \theta = a^2 - b^2 \]

Substituting \( a = 4 \) and \( b = 3 \):

\[ 4x \sec \theta - 3y \cos \theta = 16 - 9 \]

This simplifies to:

\[ 4x \sec \theta - 3y \cos \theta = 7 \]

### Step 2: Find the intersection points \( M \) and \( N \)

**Finding point \( M \) (intersection with the x-axis):**

Set \( y = 0 \):

\[ 4x \sec \theta = 7 \implies x = \frac{7}{4 \sec \theta} = \frac{7 \cos \theta}{4} \]

Thus, point \( M \) is:

\[ M = \left(\frac{7 \cos \theta}{4}, 0\right) \]

**Finding point \( N \) (intersection with the y-axis):**

Set \( x = 0 \):

\[ -3y \cos \theta = 7 \implies y = -\frac{7}{3 \sin \theta} \]

Thus, point \( N \) is:

\[ N = \left(0, -\frac{7}{3} \sin \theta\right) \]

### Step 3: Calculate the distances \( |PM| \) and \( |PN| \)

**Distance \( |PM| \):**

Using the distance formula:

\[ |PM| = \sqrt{\left(4 \cos \theta - \frac{7 \cos \theta}{4}\right)^2 + (3 \sin \theta - 0)^2} \]

Calculating \( |PM| \):

\[ |PM| = \sqrt{\left(\frac{16 \cos \theta - 7 \cos \theta}{4}\right)^2 + (3 \sin \theta)^2} \]
\[ = \sqrt{\left(\frac{9 \cos \theta}{4}\right)^2 + (3 \sin \theta)^2} \]
\[ = \sqrt{\frac{81 \cos^2 \theta}{16} + 9 \sin^2 \theta} \]
\[ = \sqrt{\frac{81 \cos^2 \theta + 144 \sin^2 \theta}{16}} = \frac{1}{4} \sqrt{81 \cos^2 \theta + 144 \sin^2 \theta} \]

**Distance \( |PN| \):**

Using the distance formula:

\[ |PN| = \sqrt{\left(4 \cos \theta - 0\right)^2 + \left(3 \sin \theta + \frac{7}{3} \sin \theta\right)^2} \]

Calculating \( |PN| \):

\[ |PN| = \sqrt{(4 \cos \theta)^2 + \left(3 \sin \theta + \frac{7}{3} \sin \theta\right)^2} \]
\[ = \sqrt{(4 \cos \theta)^2 + \left(\frac{16 \sin \theta}{3}\right)^2} \]
\[ = \sqrt{16 \cos^2 \theta + \frac{256 \sin^2 \theta}{9}} = \sqrt{\frac{144 \cos^2 \theta + 256 \sin^2 \theta}{9}} = \frac{1}{3} \sqrt{144 \cos^2 \theta + 256 \sin^2 \theta} \]

### Step 4: Find the ratio \( |PM| : |PN| \)

Now we can find the ratio:

\[ \frac{|PM|}{|PN|} = \frac{\frac{1}{4} \sqrt{81 \cos^2 \theta + 144 \sin^2 \theta}}{\frac{1}{3} \sqrt{144 \cos^2 \theta + 256 \sin^2 \theta}} \]

This simplifies to:

\[ = \frac{3}{4} \cdot \frac{\sqrt{81 \cos^2 \theta + 144 \sin^2 \theta}}{\sqrt{144 \cos^2 \theta + 256 \sin^2 \theta}} \]

After simplification, we find that:

\[ |PM| : |PN| = 9 : 16 \]

### Final Answer:

\[ |PM| : |PN| = 9 : 16 \]