Tangents are drawn from the point P(3,4) to the ellipse`x^2/9+y^2/4=1` touching the ellipse at the point A and B then the equation of the locus of the point whose distance from the point P and the line AB are equal, is

Options

A. `9x^(2)+y^(2)-6xy-54x-62y+241=0`
B. `x^(2)+9y^(2)+6xy-54x+62y-241=0`
C. `9x^(2)+9y^(2)-6xy-54x-62y-241=0`
D. `x^(2)+y^(2)-2xy+27x+31y-120=0`

Explanation

To find the equation of the locus of the point whose distance from the point P(3, 4) and the line AB (the tangents to the ellipse) are equal, we can follow these steps:

### Step 1: Write the equation of the ellipse
The given ellipse is:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]

### Step 2: Identify the point P
The point from which tangents are drawn is:
\[ P(3, 4) \]
Let \( (x_1, y_1) = (3, 4) \).

### Step 3: Write the equation of the tangent from point P
The equation of the tangent to the ellipse from point \( P(x_1, y_1) \) is given by:
\[ \frac{xx_1}{9} + \frac{yy_1}{4} = 1 \]
Substituting \( x_1 = 3 \) and \( y_1 = 4 \):
\[ \frac{3x}{9} + \frac{4y}{4} = 1 \]
This simplifies to:
\[ \frac{x}{3} + y = 1 \]
Rearranging gives:
\[ x + 3y - 3 = 0 \]

### Step 4: Find the distance from point P to the line AB
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 the line \( x + 3y - 3 = 0 \):
- \( A = 1, B = 3, C = -3 \)
- For point \( P(3, 4) \):
\[ d = \frac{|1(3) + 3(4) - 3|}{\sqrt{1^2 + 3^2}} = \frac{|3 + 12 - 3|}{\sqrt{10}} = \frac{12}{\sqrt{10}} = \frac{12\sqrt{10}}{10} = \frac{6\sqrt{10}}{5} \]

### Step 5: Set up the locus condition
Let \( Q(x, y) \) be a point such that the distance from \( P(3, 4) \) to \( Q \) is equal to the distance from \( Q \) to the line \( AB \):
\[ \sqrt{(x - 3)^2 + (y - 4)^2} = \frac{|x + 3y - 3|}{\sqrt{10}} \]

### Step 6: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (x - 3)^2 + (y - 4)^2 = \frac{(x + 3y - 3)^2}{10} \]

### Step 7: Expand both sides
Expanding the left side:
\[ (x - 3)^2 + (y - 4)^2 = (x^2 - 6x + 9) + (y^2 - 8y + 16) = x^2 + y^2 - 6x - 8y + 25 \]
Expanding the right side:
\[ \frac{(x + 3y - 3)^2}{10} = \frac{x^2 + 6xy + 9y^2 - 6x - 18y + 9}{10} \]

### Step 8: Multiply through by 10 to eliminate the fraction
\[ 10(x^2 + y^2 - 6x - 8y + 25) = x^2 + 6xy + 9y^2 - 6x - 18y + 9 \]

### Step 9: Rearrange and simplify
Bringing all terms to one side:
\[ 10x^2 + 10y^2 - 60x - 80y + 250 - x^2 - 6xy - 9y^2 + 6x + 18y - 9 = 0 \]
Combine like terms:
\[ (10 - 1)x^2 + (10 - 9)y^2 - 6xy - (60 - 6)x - (80 - 18)y + (250 - 9) = 0 \]
This simplifies to:
\[ 9x^2 + y^2 - 6xy - 54x - 62y + 241 = 0 \]

### Final Result
The equation of the locus of the point whose distance from point P and the line AB are equal is:
\[ 9x^2 + y^2 - 6xy - 54x - 62y + 241 = 0 \]