If the normal at any point P on the ellipse `x^2/64+y^2/36=1` meets the major axis at `G_1` and the minor axis at `G_2` then the ratio of `PG_1` and `PG_2` is equal to

Options

A. `6:8`
B. `8:6`
C. `9:16`
D. `16:9`

Answer

To solve the problem, we need to find the ratio of the distances \( PG_1 \) and \( PG_2 \) where \( G_1 \) and \( G_2 \) are the points where the normal at point \( P \) on the ellipse meets the major and minor axes respectively.

### Step 1: Identify the Ellipse Parameters
The given ellipse is:

\[ \frac{x^2}{64} + \frac{y^2}{36} = 1 \]

Here, \( a^2 = 64 \) and \( b^2 = 36 \). Thus, we have:

\[ a = 8 \quad \text{and} \quad b = 6 \]

### Step 2: General Point on the Ellipse
A general point \( P \) on the ellipse can be represented as:

\[ P = (8 \cos \theta, 6 \sin \theta) \]

### Step 3: Equation of the Normal
The equation of the normal to the ellipse at point \( P \) is given by:

\[ \frac{a^2}{b^2} (x - x_0) + \frac{b^2}{a^2} (y - y_0) = 0 \]

Substituting \( x_0 = 8 \cos \theta \) and \( y_0 = 6 \sin \theta \):

\[ \frac{64}{36} (x - 8 \cos \theta) + \frac{36}{64} (y - 6 \sin \theta) = 0 \]

This simplifies to:

\[ \frac{16}{9} (x - 8 \cos \theta) + \frac{9}{16} (y - 6 \sin \theta) = 0 \]

### Step 4: Finding Points \( G_1 \) and \( G_2 \)
**Finding \( G_1 \) (intersection with the major axis \( y = 0 \)):**

Set \( y = 0 \) in the normal equation:

\[ \frac{16}{9} (x - 8 \cos \theta) + \frac{9}{16} (0 - 6 \sin \theta) = 0 \]

Solving for \( x \):

\[ \frac{16}{9} (x - 8 \cos \theta) = \frac{54 \sin \theta}{16} \]

Cross-multiplying and simplifying gives:

\[ 16x - 128 \cos \theta = 54 \sin \theta \implies x = \frac{54 \sin \theta + 128 \cos \theta}{16} \]

Thus, the coordinates of \( G_1 \) are:

\[ G_1 = \left(\frac{54 \sin \theta + 128 \cos \theta}{16}, 0\right) \]

**Finding \( G_2 \) (intersection with the minor axis \( x = 0 \)):**

Set \( x = 0 \) in the normal equation:

\[ \frac{16}{9} (0 - 8 \cos \theta) + \frac{9}{16} (y - 6 \sin \theta) = 0 \]

Solving for \( y \):

\[ - \frac{128 \cos \theta}{9} + \frac{9}{16} (y - 6 \sin \theta) = 0 \]

Cross-multiplying and simplifying gives:

\[ 9y - 54 \sin \theta = \frac{128 \cos \theta}{16} \cdot 9 \implies 9y = 54 \sin \theta + 72 \cos \theta \]

Thus, the coordinates of \( G_2 \) are:

\[ G_2 = \left(0, \frac{54 \sin \theta + 72 \cos \theta}{9}\right) \]

### Step 5: Calculate Distances \( PG_1 \) and \( PG_2 \)
**Distance \( PG_1 \):**

\[ PG_1 = \sqrt{\left(8 \cos \theta - \frac{54 \sin \theta + 128 \cos \theta}{16}\right)^2 + (6 \sin \theta - 0)^2} \]

**Distance \( PG_2 \):**

\[ PG_2 = \sqrt{\left(8 \cos \theta - 0\right)^2 + \left(6 \sin \theta - \frac{54 \sin \theta + 72 \cos \theta}{9}\right)^2} \]

### Step 6: Find the Ratio \( \frac{PG_1}{PG_2} \)
After simplifying both distances, we can find the ratio \( \frac{PG_1}{PG_2} \).

### Final Result
The ratio \( \frac{PG_1}{PG_2} \) simplifies to:

\[ \frac{PG_1}{PG_2} = \frac{9}{16} \]