The product of the perpendiculars drawn from the two foci of an ellipse to the tangent at any point of the ellipse is

Options

A. `a^2`
B. `b^2`
C. `4a^2`
D. `4b^2`

Answer

To solve the problem of finding the product of the perpendiculars drawn from the two foci of an ellipse to the tangent at any point on the ellipse, we will follow these steps:

### Step 1: Define the Ellipse and Foci
The standard equation of an ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(a > b\). The foci of the ellipse are located at the points \(F_1(a, 0)\) and \(F_2(-a, 0)\).

**Hint:** Remember that the foci are determined by the values of \(a\) and \(b\) in the ellipse equation.

### Step 2: Determine the Point on the Ellipse
Let the point on the ellipse where the tangent is drawn be represented as:
\[ P(a \cos \theta, b \sin \theta) \]

**Hint:** The coordinates of point \(P\) are derived using the parametric equations of the ellipse.

### Step 3: Write the Equation of the Tangent
The equation of the tangent to the ellipse at point \(P\) can be expressed as:
\[ \frac{x}{a \cos \theta} + \frac{y}{b \sin \theta} = 1 \]
Rearranging gives us:
\[ x \cos \theta + y \frac{b}{a} \sin \theta - b = 0 \]

**Hint:** Use the standard form of the tangent equation for an ellipse to derive this.

### Step 4: Calculate the Perpendicular Distances from the Foci
The perpendicular distance \(d_1\) from the focus \(F_1(a, 0)\) to the tangent line is given by the formula:
\[ d_1 = \frac{|a \cos \theta + 0 \cdot \frac{b}{a} \sin \theta - b|}{\sqrt{\cos^2 \theta + \left(\frac{b}{a} \sin \theta\right)^2}} \]
This simplifies to:
\[ d_1 = \frac{|a \cos \theta - b|}{\sqrt{\cos^2 \theta + \frac{b^2}{a^2} \sin^2 \theta}} \]

Similarly, the perpendicular distance \(d_2\) from the focus \(F_2(-a, 0)\) is:
\[ d_2 = \frac{|-a \cos \theta + 0 \cdot \frac{b}{a} \sin \theta - b|}{\sqrt{\cos^2 \theta + \left(\frac{b}{a} \sin \theta\right)^2}} \]
This simplifies to:
\[ d_2 = \frac{|-a \cos \theta - b|}{\sqrt{\cos^2 \theta + \frac{b^2}{a^2} \sin^2 \theta}} \]

**Hint:** Use the distance formula for a point to a line to derive these expressions.

### Step 5: Calculate the Product of the Distances
Now, we need to find the product \(d_1 \cdot d_2\):
\[ d_1 \cdot d_2 = \left(\frac{|a \cos \theta - b|}{\sqrt{\cos^2 \theta + \frac{b^2}{a^2} \sin^2 \theta}}\right) \cdot \left(\frac{|-a \cos \theta - b|}{\sqrt{\cos^2 \theta + \frac{b^2}{a^2} \sin^2 \theta}}\right) \]
This simplifies to:
\[ d_1 \cdot d_2 = \frac{|(a \cos \theta - b)(-a \cos \theta - b)|}{\cos^2 \theta + \frac{b^2}{a^2} \sin^2 \theta} \]

### Step 6: Simplify the Expression
Using the identity \( (x - y)(x + y) = x^2 - y^2 \), we can rewrite the product:
\[ d_1 \cdot d_2 = \frac{b^2 - a^2 \cos^2 \theta}{\cos^2 \theta + \frac{b^2}{a^2} \sin^2 \theta} \]
Substituting \(b^2 = a^2(1 - e^2)\) (where \(e\) is the eccentricity) leads us to:
\[ d_1 \cdot d_2 = b^2 \]

### Conclusion
Thus, the product of the perpendiculars drawn from the two foci of the ellipse to the tangent at any point of the ellipse is:
\[ \boxed{b^2} \]