Tangents at the extremities of the latus rectum of an ellipse intersect on the line whose equation is .......

Answer

To find the equation of the line on which the tangents at the extremities of the latus rectum of an ellipse intersect, we can follow these steps:

### Step 1: Understand the Ellipse
We start with the standard form of the ellipse given by the equation:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(a > b\).

### Step 2: Identify the Foci and Extremities of the Latus Rectum
The foci of the ellipse are located at \((ae, 0)\) and \((-ae, 0)\), where \(e = \sqrt{1 - \frac{b^2}{a^2}}\) is the eccentricity of the ellipse. The extremities of the latus rectum are given by the points:
\[ \left(ae, \frac{2b^2}{a}\right) \quad \text{and} \quad \left(ae, -\frac{2b^2}{a}\right) \]

### Step 3: Find the Tangent Equations at the Extremities
The equation of the tangent to the ellipse at a point \((x_1, y_1)\) is:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]

#### Tangent at \(\left(ae, \frac{2b^2}{a}\right)\):
Substituting \(x_1 = ae\) and \(y_1 = \frac{2b^2}{a}\):
\[ \frac{xx_e}{a^2} + \frac{y \cdot \frac{2b^2}{a}}{b^2} = 1 \]
This simplifies to:
\[ \frac{xe}{a} + \frac{2y}{a} = 1 \quad \text{(Equation 1)} \]

#### Tangent at \(\left(ae, -\frac{2b^2}{a}\right)\):
Substituting \(x_1 = ae\) and \(y_1 = -\frac{2b^2}{a}\):
\[ \frac{xx_e}{a^2} + \frac{y \cdot -\frac{2b^2}{a}}{b^2} = 1 \]
This simplifies to:
\[ \frac{xe}{a} - \frac{2y}{a} = 1 \quad \text{(Equation 2)} \]

### Step 4: Find the Intersection of the Tangents
Now we will add Equation 1 and Equation 2:
\[ \left(\frac{xe}{a} + \frac{2y}{a}\right) + \left(\frac{xe}{a} - \frac{2y}{a}\right) = 1 + 1 \]
This simplifies to:
\[ \frac{2xe}{a} = 2 \]
Thus,
\[ x = \frac{a}{e} \]

### Step 5: Consider the Other Focus
For the other focus \((-ae, 0)\), we can repeat the process. The intersection point will yield:
\[ x = -\frac{a}{e} \]

### Step 6: Final Equation of the Line
The tangents intersect at the points \(x = \frac{a}{e}\) and \(x = -\frac{a}{e}\). Therefore, the equation of the line on which these tangents intersect is:
\[ x = \pm \frac{a}{e} \]

### Final Answer
The final answer is:
\[ x = \pm \frac{a}{e} \]