Let the two foci of an ellipse be `(-1, 0) and (3, 4)` and the foot of perpendicular from the focus `(3, 4)` upon a tangent to the ellipse be `(4, 6)`. The foot of perpendicular from the focus `(-1, 0)` upon the same tangent to the ellipse is

Options

A. `((12)/(5),(34)/(5))`
B. `((7)/(3),(11)/(3))`
C. `(2,(17)/(4))`
D. `(-1,2)`

Answer

To solve the problem, we need to find the foot of the perpendicular from the focus at (-1, 0) to the tangent line at the point (4, 6) of the ellipse.

### Step-by-Step Solution:

1. **Identify the Foci and the Point of Tangency:**
- The foci of the ellipse are given as \( F_1(-1, 0) \) and \( F_2(3, 4) \).
- The point of tangency on the ellipse is \( Q(4, 6) \).

2. **Calculate the Slope of the Line from \( F_2 \) to \( Q \):**
- The slope of the line \( F_2Q \) is calculated as:
\[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 4}{4 - 3} = \frac{2}{1} = 2 \]

3. **Determine the Slope of the Tangent Line:**
- The slope of the tangent line is the negative reciprocal of the slope of the normal line (which is the line from the focus to the point of tangency).
- Therefore, the slope of the tangent line is:
\[ m = -\frac{1}{2} \]

4. **Find the Equation of the Tangent Line:**
- Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
- Substituting \( (x_1, y_1) = (4, 6) \) and \( m = -\frac{1}{2} \):
\[ y - 6 = -\frac{1}{2}(x - 4) \]
- Simplifying this, we get:
\[ y - 6 = -\frac{1}{2}x + 2 \implies y = -\frac{1}{2}x + 8 \]
- Rearranging gives us the equation of the tangent:
\[ x + 2y = 16 \]

5. **Set Up the Equation for the Foot of the Perpendicular from \( F_1 \):**
- Let the foot of the perpendicular from \( F_1 \) be \( P(h, k) \).
- Since \( P \) lies on the tangent line, we have:
\[ h + 2k = 16 \quad \text{(Equation 1)} \]

6. **Find the Slope of the Line from \( F_1 \) to \( P \):**
- The slope of the line \( F_1P \) is:
\[ \text{slope} = \frac{k - 0}{h + 1} \]
- This slope should equal the slope of the normal line, which is \( 2 \):
\[ \frac{k}{h + 1} = 2 \implies k = 2(h + 1) \quad \text{(Equation 2)} \]

7. **Substitute Equation 2 into Equation 1:**
- Substitute \( k = 2(h + 1) \) into \( h + 2k = 16 \):
\[ h + 2(2(h + 1)) = 16 \]
- Simplifying gives:
\[ h + 4h + 4 = 16 \implies 5h + 4 = 16 \implies 5h = 12 \implies h = \frac{12}{5} \]

8. **Find \( k \) Using the Value of \( h \):**
- Substitute \( h = \frac{12}{5} \) back into Equation 2:
\[ k = 2\left(\frac{12}{5} + 1\right) = 2\left(\frac{12}{5} + \frac{5}{5}\right) = 2\left(\frac{17}{5}\right) = \frac{34}{5} \]

9. **Conclusion:**
- The foot of the perpendicular from the focus \( (-1, 0) \) upon the tangent to the ellipse is:
\[ P\left(\frac{12}{5}, \frac{34}{5}\right) \]