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 equation of the locus of the points 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`

Answer

To solve the problem of finding the locus of points whose distance from point P(3, 4) and the line AB (the chord of contact) are equal, we will follow these steps:

### Step 1: Identify the equation of the ellipse and the point P.
The given ellipse is:
\[ \frac{x^2}{9} + \frac{y^2}{4} = 1 \]
This implies \( a^2 = 9 \) and \( b^2 = 4 \). The point from which tangents are drawn is \( P(3, 4) \).

### Step 2: Write the equation of the chord of contact.
The equation of the chord of contact from point \( P(x_1, y_1) \) to the ellipse is given by:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]
Substituting \( x_1 = 3 \) and \( y_1 = 4 \), we have:
\[ \frac{3x}{9} + \frac{4y}{4} = 1 \]
Simplifying this gives:
\[ \frac{x}{3} + y = 1 \quad \Rightarrow \quad x + 3y = 3 \]
This is the equation of the chord of contact (line AB).

### Step 3: Find the distance from point P to line AB.
The distance \( d \) from point \( P(3, 4) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( x + 3y - 3 = 0 \), we have \( A = 1, B = 3, C = -3 \). Thus, the distance from \( P(3, 4) \) is:
\[ d = \frac{|1(3) + 3(4) - 3|}{\sqrt{1^2 + 3^2}} = \frac{|3 + 12 - 3|}{\sqrt{1 + 9}} = \frac{12}{\sqrt{10}} = \frac{12\sqrt{10}}{10} = \frac{6\sqrt{10}}{5} \]

### Step 4: Set up the equation for the locus of point Q.
Let \( Q(h, k) \) be a point such that the distance from \( Q \) to \( P \) is equal to the distance from \( Q \) to the line \( AB \).

1. Distance from \( Q(h, k) \) to \( P(3, 4) \):
\[ d_1 = \sqrt{(h - 3)^2 + (k - 4)^2} \]

2. Distance from \( Q(h, k) \) to the line \( x + 3y - 3 = 0 \):
\[ d_2 = \frac{|h + 3k - 3|}{\sqrt{1^2 + 3^2}} = \frac{|h + 3k - 3|}{\sqrt{10}} \]

### Step 5: Set the distances equal and square both sides.
Setting \( d_1 = d_2 \):
\[ \sqrt{(h - 3)^2 + (k - 4)^2} = \frac{|h + 3k - 3|}{\sqrt{10}} \]
Squaring both sides:
\[ (h - 3)^2 + (k - 4)^2 = \frac{(h + 3k - 3)^2}{10} \]

### Step 6: Expand and simplify the equation.
1. Expanding the left side:
\[ (h - 3)^2 + (k - 4)^2 = (h^2 - 6h + 9) + (k^2 - 8k + 16) = h^2 + k^2 - 6h - 8k + 25 \]

2. 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 and rearrange.
Multiplying through by 10 gives:
\[ 10(h^2 + k^2 - 6h - 8k + 25) = h^2 + 6hk + 9k^2 - 6h - 18k + 9 \]
Rearranging leads to:
\[ 9h^2 + k^2 - 6hk - 54h - 62k + 241 = 0 \]

### Step 8: Replace \( h \) and \( k \) with \( x \) and \( y \).
Finally, replacing \( h \) and \( k \) with \( x \) and \( y \), we have:
\[ 9x^2 + y^2 - 6xy - 54x - 62y + 241 = 0 \]

This is the required equation of the locus of point \( Q \).