If the loucus of the feet of perpendicular from the foci on any tangent to an ellipse `(x^(2))/(4) + (y^(2))/(2) =1` is `x^(2) + y^(2) =k`, then the value of k is _______.

Answer

To solve the problem, we need to find the value of \( k \) such that the locus of the feet of the perpendiculars from the foci of the ellipse to any tangent is given by the equation \( x^2 + y^2 = k \).

**Step 1: Identify the ellipse parameters.**
The given ellipse is:
\[ \frac{x^2}{4} + \frac{y^2}{2} = 1 \]
From this equation, we can identify \( a^2 = 4 \) and \( b^2 = 2 \). Thus, \( a = 2 \) and \( b = \sqrt{2} \).

**Step 2: Calculate the eccentricity \( e \) of the ellipse.**
The eccentricity \( e \) of the ellipse is given by:
\[ e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{2}{4}} = \sqrt{1 - \frac{1}{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

**Step 3: Determine the foci of the ellipse.**
The foci of the ellipse are located at:
\[ (\pm ae, 0) = \left(\pm 2 \cdot \frac{1}{\sqrt{2}}, 0\right) = \left(\pm \sqrt{2}, 0\right) \]

**Step 4: Write the equation of the tangent line.**
The equation of the tangent to the ellipse at a point \( P(x_1, y_1) \) can be written as:
\[ \frac{x x_1}{4} + \frac{y y_1}{2} = 1 \]
For a point \( P \) on the ellipse, we can express \( P \) in terms of the angle \( \theta \):
\[ x_1 = 2 \cos \theta, \quad y_1 = \sqrt{2} \sin \theta \]
Thus, the equation of the tangent becomes:
\[ \frac{x (2 \cos \theta)}{4} + \frac{y (\sqrt{2} \sin \theta)}{2} = 1 \]
Simplifying this gives:
\[ \frac{x \cos \theta}{2} + y \sin \theta = 1 \]

**Step 5: Find the slope of the tangent line.**
The slope \( m_1 \) of the tangent line can be derived from the equation:
\[ y = -\frac{\cos \theta}{2 \sin \theta} x + \frac{1}{\sin \theta} \]
Thus, the slope \( m_1 \) is:
\[ m_1 = -\frac{\cos \theta}{2 \sin \theta} \]

**Step 6: Find the equation of the perpendicular from the focus.**
The slope of the perpendicular from the focus \( F(\sqrt{2}, 0) \) will be:
\[ m_2 = \frac{2 \sin \theta}{\cos \theta} \]
The equation of the line passing through the focus can be written as:
\[ y - 0 = m_2 (x - \sqrt{2}) \]
This simplifies to:
\[ y = \frac{2 \sin \theta}{\cos \theta} (x - \sqrt{2}) \]

**Step 7: Find the locus of the feet of the perpendiculars.**
To find the locus, we need to eliminate \( \theta \) from the equations. After some algebraic manipulations involving the tangent and the perpendicular, we arrive at:
\[ x^2 + y^2 = 4 \]

**Step 8: Identify the value of \( k \).**
From the equation \( x^2 + y^2 = k \), we can see that:
\[ k = 4 \]

Thus, the value of \( k \) is \( \boxed{4} \).

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