The line `2x+y=3` intersects the ellipse `4x^(2)+y^(2)=5` at two points. The point of intersection of the tangents to the ellipse at these point is

Options

A. `((5)/(6),(5)/(6))`
B. `((5)/(3),(5)/(6))`
C. `((5)/(6),(5)/(3))`
D. `((5)/(3),(5)/(3))`

Answer

To find the point of intersection of the tangents to the ellipse at the points where the line intersects the ellipse, we can follow these steps:

### Step 1: Write the equations of the line and the ellipse.
The line is given by:
\[ 2x + y = 3 \]
The ellipse is given by:
\[ 4x^2 + y^2 = 5 \]

### Step 2: Substitute the line equation into the ellipse equation.
From the line equation, we can express \( y \) in terms of \( x \):
\[ y = 3 - 2x \]
Now, substitute this expression for \( y \) into the ellipse equation:
\[ 4x^2 + (3 - 2x)^2 = 5 \]

### Step 3: Expand and simplify the equation.
Expanding the equation:
\[ 4x^2 + (3 - 2x)(3 - 2x) = 5 \]
\[ 4x^2 + (9 - 12x + 4x^2) = 5 \]
Combining like terms:
\[ 8x^2 - 12x + 9 = 5 \]
Subtracting 5 from both sides:
\[ 8x^2 - 12x + 4 = 0 \]

### Step 4: Solve the quadratic equation.
Now we can simplify the quadratic equation:
\[ 2x^2 - 3x + 1 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = -3, c = 1 \):
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \]
\[ x = \frac{3 \pm \sqrt{9 - 8}}{4} \]
\[ x = \frac{3 \pm 1}{4} \]
This gives us two solutions:
\[ x_1 = 1 \quad \text{and} \quad x_2 = \frac{1}{2} \]

### Step 5: Find the corresponding \( y \) values.
Using the line equation \( y = 3 - 2x \):
For \( x_1 = 1 \):
\[ y_1 = 3 - 2(1) = 1 \]
For \( x_2 = \frac{1}{2} \):
\[ y_2 = 3 - 2\left(\frac{1}{2}\right) = 2 \]

### Step 6: Find the point of intersection of the tangents.
The points of intersection on the ellipse are \( (1, 1) \) and \( \left(\frac{1}{2}, 2\right) \).

Using the formula for the point of intersection of tangents at points \( (x_1, y_1) \) and \( (x_2, y_2) \):
Let \( (x_1, y_1) = (1, 1) \) and \( (x_2, y_2) = \left(\frac{1}{2}, 2\right) \):
The coordinates of the point of intersection \( (h, k) \) can be found using:
\[ h = \frac{x_1 + x_2}{2} = \frac{1 + \frac{1}{2}}{2} = \frac{3}{4} \]
\[ k = \frac{y_1 + y_2}{2} = \frac{1 + 2}{2} = \frac{3}{2} \]

Thus, the point of intersection of the tangents is:
\[ \left(\frac{3}{4}, \frac{3}{2}\right) \]