Tangents are drawn from the point P(3,4) to the ellipse `x^(2)/9+y^(2)/4=1` touching the ellipse at point A and B. Q. The coordinates of A and B are

Options

A. (3,0) amd (0,2)
B. `(-8/5,(2sqrt161/15)and (-9/8,8/5)`
C. `(-8/5,(2sqrt161/15)and(0,2)`
D. `(3,0) and (-9/5,8/5)`

Answer

To find the coordinates of the points A and B where the tangents from the point P(3, 4) touch the ellipse given by the equation \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \), we will follow these steps:

### Step 1: Identify the equation of the ellipse
The equation of the ellipse is given as:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
Here, \( a^2 = 9 \) (so \( a = 3 \)) and \( b^2 = 4 \) (so \( b = 2 \)).

### Step 2: Use the formula for the chord of contact
The chord of contact from a point \( (x_1, y_1) \) to the ellipse is given by:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]
Substituting \( (x_1, y_1) = (3, 4) \), \( a^2 = 9 \), and \( b^2 = 4 \):
\[ \frac{3x}{9} + \frac{4y}{4} = 1 \]
This simplifies to:
\[ \frac{x}{3} + y = 1 \]
or
\[ x + 3y = 3 \quad \text{(Equation 1)} \]

### Step 3: Substitute Equation 1 into the ellipse equation
Now, we will substitute \( x \) from Equation 1 into the ellipse equation. From Equation 1, we can express \( x \) as:
\[ x = 3 - 3y \]
Substituting this into the ellipse equation:
\[ \frac{(3 - 3y)^2}{9} + \frac{y^2}{4} = 1 \]
Expanding this:
\[ \frac{(9 - 18y + 9y^2)}{9} + \frac{y^2}{4} = 1 \]
Simplifying:
\[ 1 - 2y + y^2 + \frac{y^2}{4} = 1 \]
Multiplying through by 36 to eliminate the denominators:
\[ 36(1 - 2y + y^2) + 9y^2 = 36 \]
This simplifies to:
\[ 36 - 72y + 36y^2 + 9y^2 = 36 \]
Combining like terms:
\[ 45y^2 - 72y = 0 \]

### Step 4: Factor the equation
Factoring out \( y \):
\[ y(45y - 72) = 0 \]
Setting each factor to zero gives:
1. \( y = 0 \)
2. \( 45y - 72 = 0 \) → \( y = \frac{72}{45} = \frac{8}{5} \)

### Step 5: Find corresponding x-coordinates
Now, we will find the corresponding x-coordinates for each y-value.

1. For \( y = 0 \):
\[ x = 3 - 3(0) = 3 \]
So, one point is \( A(3, 0) \).

2. For \( y = \frac{8}{5} \):
\[ x = 3 - 3\left(\frac{8}{5}\right) = 3 - \frac{24}{5} = \frac{15}{5} - \frac{24}{5} = -\frac{9}{5} \]
So, the second point is \( B\left(-\frac{9}{5}, \frac{8}{5}\right) \).

### Final Answer
The coordinates of points A and B are:
- A: \( (3, 0) \)
- B: \( \left(-\frac{9}{5}, \frac{8}{5}\right) \)