If foci of an ellipse are `(-2,3),(5,9)` & `2x+3y+15=0` is tangent to the ellipse. Then find the point of contact the tangent cannot be

Options

A. `((7)/(2),-1)`
B. `(-(9)/(2),-2)`
C. `(-1,(9)/(2))`
D. `(-1,(7)/(2))`

Explanation

To solve the problem, we need to find the point of contact where the tangent to the ellipse cannot be, given the foci and the tangent line equation. Here’s a step-by-step solution:

### Step 1: Find the center of the ellipse
The foci of the ellipse are given as \( F_1(-2, 3) \) and \( F_2(5, 9) \). The center of the ellipse can be found by calculating the midpoint of the line segment joining the foci.

\[ \text{Center} (h, k) = \left( \frac{-2 + 5}{2}, \frac{3 + 9}{2} \right) = \left( \frac{3}{2}, 6 \right) \]

### Step 2: Find the distance between the foci
The distance \( 2c \) between the foci is given by the distance formula:

\[ 2c = \sqrt{(5 - (-2))^2 + (9 - 3)^2} = \sqrt{(5 + 2)^2 + (9 - 3)^2} = \sqrt{7^2 + 6^2} = \sqrt{49 + 36} = \sqrt{85} \]

Thus, \( c = \frac{\sqrt{85}}{2} \).

### Step 3: Find the slope of the tangent line
The equation of the tangent line is given as \( 2x + 3y + 15 = 0 \). We can rewrite it in slope-intercept form \( y = mx + b \):

\[ 3y = -2x - 15 \implies y = -\frac{2}{3}x - 5 \]

The slope \( m \) of the tangent line is \( -\frac{2}{3} \).

### Step 4: Find the distance from the center to the tangent line
The distance \( d \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:

\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

For our tangent line \( 2x + 3y + 15 = 0 \), we have \( A = 2, B = 3, C = 15 \) and the center \( (h, k) = \left( \frac{3}{2}, 6 \right) \):

\[ d = \frac{|2 \cdot \frac{3}{2} + 3 \cdot 6 + 15|}{\sqrt{2^2 + 3^2}} = \frac{|3 + 18 + 15|}{\sqrt{4 + 9}} = \frac{|36|}{\sqrt{13}} = \frac{36}{\sqrt{13}} \]

### Step 5: Find the semi-major axis \( a \) and semi-minor axis \( b \)
Using the relationship \( c^2 = a^2 - b^2 \) and knowing that the distance from the center to the tangent line must equal \( \frac{b}{a} \):

Let \( a \) be the semi-major axis and \( b \) be the semi-minor axis. We can use the distance calculated to find \( a \) and \( b \).

### Step 6: Determine the point of contact
To find the point of contact where the tangent cannot be, we need to check the points on the ellipse and see which ones do not satisfy the tangent condition. This requires substituting the coordinates of potential points into the tangent line equation to check if they satisfy it.

### Conclusion
The point of contact cannot be determined without additional information about the ellipse's dimensions (i.e., \( a \) and \( b \)). However, the steps outlined provide a method to find the point of contact based on the conditions given.