The length ofthe normal (terminated by the major axis) at a point of the ellipse `x^2/a^2+y^2/b^2=1` is

Options

A. `(b)/(a)(r+r_1)`
B. `(b)/(a)|r-r_1|`
C. `(b)/(a)sqrt(r r_1)`
D. independent of `r, r_1`

Answer

To find the length of the normal terminated by the major axis at a point of the ellipse given by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Ellipse**:
The standard form of the ellipse is given by \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \( a \) is the semi-major axis and \( b \) is the semi-minor axis. The foci of the ellipse are located at \( (c, 0) \) and \( (-c, 0) \) where \( c = \sqrt{a^2 - b^2} \).

2. **Point on the Ellipse**:
Let \( P(x_0, y_0) \) be a point on the ellipse. The coordinates of this point satisfy the ellipse equation.

3. **Focal Distances**:
The distances from point \( P \) to the foci \( F_1(c, 0) \) and \( F_2(-c, 0) \) are denoted as \( r \) and \( r_1 \) respectively. By the property of ellipses, we know that:
\[ r + r_1 = 2a \]

4. **Normal to the Ellipse**:
The normal at point \( P \) intersects the major axis at point \( D \). The lengths \( F_1D \) and \( F_2D \) can be expressed in terms of \( r \) and \( r_1 \):
\[ \frac{F_1D}{F_2D} = \frac{r}{r_1} \]

5. **Using the Angle Bisector Theorem**:
From the angle bisector theorem, we can express the distances:
\[ F_1D = \frac{r}{r + r_1} \cdot F_1F_2 \quad \text{and} \quad F_2D = \frac{r_1}{r + r_1} \cdot F_1F_2 \]
Since \( F_1F_2 = 2c \), we can write:
\[ F_1D = \frac{r \cdot 2c}{2a} = \frac{rc}{a} \quad \text{and} \quad F_2D = \frac{r_1 \cdot 2c}{2a} = \frac{r_1c}{a} \]

6. **Length of the Normal**:
The length of the normal \( PD \) can be calculated using the relationship derived from the cosine rule or by using the properties of the ellipse. The length \( PD \) is given by:
\[ PD = \sqrt{r r_1 \left(1 - e^2\right)} \]
where \( e = \frac{c}{a} = \sqrt{1 - \frac{b^2}{a^2}} \).

7. **Substituting for \( e^2 \)**:
We know \( e^2 = 1 - \frac{b^2}{a^2} \), so:
\[ 1 - e^2 = \frac{b^2}{a^2} \]
Therefore, substituting this back, we get:
\[ PD = \sqrt{r r_1 \cdot \frac{b^2}{a^2}} = \frac{b}{a} \sqrt{r r_1} \]

### Final Result:
The length of the normal terminated by the major axis at a point of the ellipse is:
\[ \boxed{\frac{b}{a} \sqrt{r r_1}} \]