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

Options

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

Answer

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

### Step 1: Write the equation of the chord of contact
The equation of the chord of contact from the point \( P(x_1, y_1) = (3, 4) \) to the ellipse is given by:

\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]

Here, \( a^2 = 9 \) and \( b^2 = 4 \). Substituting the values, we get:

\[ \frac{3x}{9} + \frac{4y}{4} = 1 \]

Simplifying this, we have:

\[ \frac{x}{3} + y = 1 \]

Rearranging gives us the equation of the chord of contact:

\[ x + 3y - 3 = 0 \]

### Step 2: Substitute the chord of contact equation into the ellipse equation
Next, we substitute \( y \) from the chord of contact equation into the ellipse equation. From the chord of contact, we have:

\[ y = 1 - \frac{x}{3} \]

Substituting this into the ellipse equation:

\[ \frac{x^2}{9} + \frac{(1 - \frac{x}{3})^2}{4} = 1 \]

### Step 3: Solve the equation
Expanding the second term:

\[ (1 - \frac{x}{3})^2 = 1 - \frac{2x}{3} + \frac{x^2}{9} \]

Thus, the ellipse equation becomes:

\[ \frac{x^2}{9} + \frac{1 - \frac{2x}{3} + \frac{x^2}{9}}{4} = 1 \]

Multiplying through by 36 to eliminate the denominators:

\[ 4x^2 + 9(1 - \frac{2x}{3} + \frac{x^2}{9}) = 36 \]

Expanding gives:

\[ 4x^2 + 9 - 6x + x^2 = 36 \]

Combining like terms:

\[ 5x^2 - 6x - 27 = 0 \]

### Step 4: Use the quadratic formula
Now we can apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 5, b = -6, c = -27 \):

\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 5 \cdot (-27)}}{2 \cdot 5} \]

Calculating the discriminant:

\[ x = \frac{6 \pm \sqrt{36 + 540}}{10} = \frac{6 \pm \sqrt{576}}{10} = \frac{6 \pm 24}{10} \]

This gives us two values for \( x \):

\[ x_1 = \frac{30}{10} = 3, \quad x_2 = \frac{-18}{10} = -1.8 \]

### Step 5: Find corresponding \( y \) values
Now we substitute these \( x \) values back into the equation of the chord of contact to find \( y \):

For \( x = 3 \):

\[ y = 1 - \frac{3}{3} = 0 \quad \Rightarrow \quad A(3, 0) \]

For \( x = -1.8 \):

\[ y = 1 - \frac{-1.8}{3} = 1 + 0.6 = 1.6 \quad \Rightarrow \quad B(-1.8, 1.6) \]

### Final Coordinates
Thus, the coordinates of points A and B are:

- \( A(3, 0) \)
- \( B(-1.8, 1.6) \)