The user wants to find the equation of the locus of a point that is equidistant from a given point P and a given line AB.

Step 1: Identify the given information

• The ellipse is given by the equation: $\frac{x^2}{9} + \frac{y^2}{4} = 1$ Here, $a^2 = 9$ and $b^2 = 4$.
• The external point from which tangents are drawn is $P(3, 4)$.
• The tangents from P touch the ellipse at points A and B.
• The line segment AB is the chord of contact.

Step 2: Find the equation of the chord of contact (line AB)

The equation of the chord of contact of tangents drawn from an external point $(x_1, y_1)$ to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by the formula $T=0$, which is: $\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1$ Substituting the coordinates of point $P(x_1, y_1) = (3, 4)$ and the values of $a^2=9$ and $b^2=4$ into this formula: $\frac{x(3)}{9} + \frac{y(4)}{4} = 1$ $\frac{x}{3} + y = 1$ To clear the fraction, we can multiply the entire equation by 3: $x + 3y = 3$ Rearranging this into the standard form $Ax + By + C = 0$, we get: $x + 3y - 3 = 0$ This is the equation of the line AB.

Step 3: Set up the locus condition

Let the point on the locus be $M(x, y)$. The problem states that the locus of point M is such that its distance from point P is equal to its perpendicular distance from the line AB. This is the definition of a parabola, where P is the focus and the line AB is the directrix.

The condition is: Distance(M, P) = Perpendicular distance(M, line AB).

Step 4: Calculate the distances

1. Distance between M(x, y) and P(3, 4): Using the distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$: $MP = \sqrt{(x-3)^2 + (y-4)^2}$

2. Perpendicular distance from M(x, y) to the line AB ($x + 3y - 3 = 0$): Using the formula for the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$, which is $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$: $\text{Distance} = \frac{|x + 3y - 3|}{\sqrt{1^2 + 3^2}} = \frac{|x + 3y - 3|}{\sqrt{1 + 9}} = \frac{|x + 3y - 3|}{\sqrt{10}}$

Step 5: Form the equation of the locus

Equating the two distances: $\sqrt{(x-3)^2 + (y-4)^2} = \frac{|x + 3y - 3|}{\sqrt{10}}$ To eliminate the square root and the absolute value, we square both sides of the equation: $(x-3)^2 + (y-4)^2 = \frac{(x + 3y - 3)^2}{10}$ $10[(x-3)^2 + (y-4)^2] = (x + 3y - 3)^2$

Step 6: Expand and simplify the equation

Expand the terms on both sides: Left side: $10[(x^2 - 6x + 9) + (y^2 - 8y + 16)] = 10(x^2 + y^2 - 6x - 8y + 25)$ $= 10x^2 + 10y^2 - 60x - 80y + 250$ Right side (using $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$): $(x + 3y - 3)^2 = x^2 + (3y)^2 + (-3)^2 + 2(x)(3y) + 2(3y)(-3) + 2(-3)(x)$ $= x^2 + 9y^2 + 9 + 6xy - 18y - 6x$ Now, equate the expanded forms: $10x^2 + 10y^2 - 60x - 80y + 250 = x^2 + 9y^2 + 6xy - 6x - 18y + 9$ Move all terms to one side to set the equation to zero: $(10x^2 - x^2) + (10y^2 - 9y^2) - 6xy + (-60x + 6x) + (-80y + 18y) + (250 - 9) = 0$ $9x^2 + y^2 - 6xy - 54x - 62y + 241 = 0$

Step 7: Compare with the options

The derived equation is $9x^2 + y^2 - 6xy - 54x - 62y + 241 = 0$. This matches option A.

Therefore, the correct equation of the locus is $9{x^2} + {y^2} - 6xy - 54x - 62y + 241 = 0$.