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 distances from the point P and the line AB are equal, is

Options

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

Answer

To find the equation of the locus of the point whose distances from the point P(3, 4) and the line AB are equal, we will follow these steps:

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

### Step 2: Find the equation of the chord of contact
The chord of contact from the point P(3, 4) to the ellipse can be derived using the formula:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]
where \( (x_1, y_1) \) is the point from which the tangents are drawn, and \( a^2 = 9 \) and \( b^2 = 4 \) are the semi-major and semi-minor axes respectively.

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:
\[ x + 3y - 3 = 0 \]
This is the equation of the line AB.

### Step 3: Set up the distance condition
Let \( (h, k) \) be the point whose distances from P(3, 4) and the line AB are equal. The distance from point P to point \( (h, k) \) is given by:
\[ d_1 = \sqrt{(h - 3)^2 + (k - 4)^2} \]

The distance from point \( (h, k) \) to the line \( x + 3y - 3 = 0 \) can be calculated using the formula:
\[ d_2 = \frac{|h + 3k - 3|}{\sqrt{1^2 + 3^2}} = \frac{|h + 3k - 3|}{\sqrt{10}} \]

### Step 4: Set the distances equal
Setting \( d_1 \) equal to \( d_2 \):
\[ \sqrt{(h - 3)^2 + (k - 4)^2} = \frac{|h + 3k - 3|}{\sqrt{10}} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (h - 3)^2 + (k - 4)^2 = \frac{(h + 3k - 3)^2}{10} \]

### Step 6: Expand both sides
Expanding the left side:
\[ (h - 3)^2 + (k - 4)^2 = (h^2 - 6h + 9) + (k^2 - 8k + 16) = h^2 + k^2 - 6h - 8k + 25 \]

Expanding the right side:
\[ \frac{(h + 3k - 3)^2}{10} = \frac{(h^2 + 6hk + 9k^2 - 6h - 18k + 9)}{10} \]

### Step 7: Multiply through by 10 to eliminate the fraction
Multiplying through by 10 gives:
\[ 10(h^2 + k^2 - 6h - 8k + 25) = h^2 + 6hk + 9k^2 - 6h - 18k + 9 \]

### Step 8: Rearranging the equation
Rearranging all terms to one side results in:
\[ 10h^2 + 10k^2 - 60h - 80k + 250 - h^2 - 6hk - 9k^2 + 6h + 18k - 9 = 0 \]
Combining like terms gives:
\[ 9h^2 + k^2 - 6hk - 54h - 62k + 241 = 0 \]

### Step 9: Replace \( h \) and \( k \) with \( x \) and \( y \)
Thus, the final equation of the locus is:
\[ 9x^2 + y^2 - 6xy - 54x - 62y + 241 = 0 \]