To find the equations of the tangent and normal at the ends of the latus rectum of the ellipse given by \( 9x^2 + 16y^2 = 144 \), we will follow these steps:
### Step 1: Convert the equation of the ellipse to standard form
The given equation is:
\[
9x^2 + 16y^2 = 144
\]
Dividing the entire equation by 144, we get:
\[
\frac{x^2}{16} + \frac{y^2}{9} = 1
\]
This is the standard form of the ellipse, where \( a^2 = 16 \) and \( b^2 = 9 \). Thus, \( a = 4 \) and \( b = 3 \).
### Step 2: Find the coordinates of the ends of the latus rectum
The latus rectum of an ellipse is given by the formula:
\[
\left( \frac{b^2}{a}, \pm a \right)
\]
Here, \( b^2 = 9 \) and \( a = 4 \). Therefore, the coordinates of the ends of the latus rectum are:
\[
\left( \frac{9}{4}, 3 \right) \quad \text{and} \quad \left( \frac{9}{4}, -3 \right)
\]
### Step 3: Find the equation of the tangent at the ends of the latus rectum
The general equation of the tangent to the ellipse at the point \( (x_1, y_1) \) is given by:
\[
\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1
\]
#### For the point \( \left( \frac{9}{4}, 3 \right) \):
Substituting \( x_1 = \frac{9}{4} \) and \( y_1 = 3 \):
\[
\frac{x \cdot \frac{9}{4}}{16} + \frac{y \cdot 3}{9} = 1
\]
Multiplying through by 144 to eliminate the denominators:
\[
9x + 48y = 144
\]
Simplifying gives:
\[
3x + 16y = 48 \quad \text{(Equation 1)}
\]
#### For the point \( \left( \frac{9}{4}, -3 \right) \):
Using the same method:
\[
\frac{x \cdot \frac{9}{4}}{16} + \frac{y \cdot (-3)}{9} = 1
\]
Multiplying through by 144:
\[
9x - 48y = 144
\]
Simplifying gives:
\[
3x - 16y = 48 \quad \text{(Equation 2)}
\]
### Step 4: Find the equations of the normals at the ends of the latus rectum
The general equation of the normal to the ellipse at the point \( (x_1, y_1) \) is given by:
\[
\frac{a^2 (x - x_1)}{x_1^2} + \frac{b^2 (y - y_1)}{y_1^2} = 0
\]
#### For the point \( \left( \frac{9}{4}, 3 \right) \):
Substituting:
\[
\frac{16(x - \frac{9}{4})}{\left(\frac{9}{4}\right)^2} + \frac{9(y - 3)}{3^2} = 0
\]
Calculating \( \left(\frac{9}{4}\right)^2 = \frac{81}{16} \):
\[
\frac{16(x - \frac{9}{4})}{\frac{81}{16}} + \frac{9(y - 3)}{9} = 0
\]
This simplifies to:
\[
\frac{256(x - \frac{9}{4})}{81} + (y - 3) = 0
\]
Rearranging gives:
\[
256x - 576 + 81y = 0 \quad \text{(Equation 3)}
\]
#### For the point \( \left( \frac{9}{4}, -3 \right) \):
Using the same method:
\[
\frac{16(x - \frac{9}{4})}{\left(\frac{9}{4}\right)^2} + \frac{9(y + 3)}{9} = 0
\]
This simplifies to:
\[
256x - 576 + 81y + 243 = 0
\]
Rearranging gives:
\[
256x + 81y - 333 = 0 \quad \text{(Equation 4)}
\]
### Summary of Results
1. **Tangent Equations**:
- At \( \left( \frac{9}{4}, 3 \right) \): \( 3x + 16y = 48 \)
- At \( \left( \frac{9}{4}, -3 \right) \): \( 3x - 16y = 48 \)
2. **Normal Equations**:
- At \( \left( \frac{9}{4}, 3 \right) \): \( 256x + 81y = 576 \)
- At \( \left( \frac{9}{4}, -3 \right) \): \( 256x + 81y = 333 \)
