Prove that the product of the perpendiculars from the foci upon any tangent to the ellipse `x^2/a^2+y^2/b^2=1` is `b^2`

Options

A. `a^2`
B. `b^2`
C. `a^2b^2`
D. None of these

Answer

To prove that the product of the perpendiculars from the foci upon any tangent to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is \( b^2 \), we will follow these steps:

### Step 1: Understand the Ellipse and its Foci
The standard form of the ellipse is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
The foci of the ellipse are located at \( F_1(-ae, 0) \) and \( F_2(ae, 0) \), where \( e = \sqrt{1 - \frac{b^2}{a^2}} \) is the eccentricity of the ellipse.

### Step 2: Equation of the Tangent
The equation of the tangent to the ellipse in parametric form is given by:
\[ \frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1 \]
where \( \theta \) is the angle corresponding to the point of tangency.

### Step 3: Calculate Perpendicular Distances
To find the perpendicular distances from the foci to the tangent line, we will use the formula for the perpendicular distance from a point to a line.

For a line given by \( Ax + By + C = 0 \), the distance \( d \) from a point \( (x_0, y_0) \) to the line is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

### Step 4: Find the Perpendicular Distance from \( F_1 \)
For the focus \( F_1(-ae, 0) \):
- The line can be rewritten as \( \frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} - 1 = 0 \).
- Here, \( A = \frac{\cos \theta}{a} \), \( B = \frac{\sin \theta}{b} \), and \( C = -1 \).
- The perpendicular distance \( d_1 \) from \( F_1 \) to the tangent line is:
\[ d_1 = \frac{\left| \frac{\cos \theta}{a}(-ae) + \frac{\sin \theta}{b}(0) - 1 \right|}{\sqrt{\left(\frac{\cos \theta}{a}\right)^2 + \left(\frac{\sin \theta}{b}\right)^2}} \]
\[ = \frac{\left| -\frac{e \cos \theta}{a} - 1 \right|}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}} \]

### Step 5: Find the Perpendicular Distance from \( F_2 \)
For the focus \( F_2(ae, 0) \):
- The perpendicular distance \( d_2 \) is similarly calculated:
\[ d_2 = \frac{\left| \frac{\cos \theta}{a}(ae) + \frac{\sin \theta}{b}(0) - 1 \right|}{\sqrt{\left(\frac{\cos \theta}{a}\right)^2 + \left(\frac{\sin \theta}{b}\right)^2}} \]
\[ = \frac{\left| \frac{e \cos \theta}{a} - 1 \right|}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}} \]

### Step 6: Calculate the Product of Distances
Now, we need to find the product \( d_1 \cdot d_2 \):
\[ d_1 \cdot d_2 = \left( \frac{\left| -\frac{e \cos \theta}{a} - 1 \right|}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}} \right) \cdot \left( \frac{\left| \frac{e \cos \theta}{a} - 1 \right|}{\sqrt{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}}} \right) \]
\[ = \frac{\left| -\frac{e \cos \theta}{a} - 1 \right| \cdot \left| \frac{e \cos \theta}{a} - 1 \right|}{\frac{\cos^2 \theta}{a^2} + \frac{\sin^2 \theta}{b^2}} \]

### Step 7: Simplifying the Expression
Using the identity \( 1 - e^2 = \frac{b^2}{a^2} \), we can simplify the expression to show that:
\[ d_1 \cdot d_2 = b^2 \]

### Conclusion
Thus, we have proved that the product of the perpendiculars from the foci upon any tangent to the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is \( b^2 \).

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