AB is variable line sliding between the coordinate axes in such a way that A lies on x-axis and B lies on y-axis. If P is a variable point on AB such that PA=b, PB=a and AB=a+b, then the equation of the locus of P is

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$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

$x^{2} - y^{2} = a^{2} - b^{2}$

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$x^{2} + y^{2} = a^{2} + b^{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} L e t A (α, 0) B (0, β) \\ G i v e n P A : P B = b : a = m : n \end{array}$

Let P ( x, y ) be the point on locus

$\begin{array}{l} P (x, y) = (\frac{b (0) + a (α)}{b + a}, \frac{b (β) + a (0)}{b + a}) \\ x = \frac{a α}{b + a}, y = \frac{b β}{a + b} \\ α = \frac{(b + a) x}{a}, β = \frac{(b + a) y}{b} \\ G i v e n A B = a + b \\ \Rightarrow \sqrt{α^{2} + β^{2}} = a + b \\ \Rightarrow α^{2} + β^{2} = {(a + b)}^{2} \\ \Rightarrow {(a + b)}^{2} \frac{x^{2}}{a^{2}} + {(a + b)}^{2} \frac{y^{2}}{b^{2}} = {(a + b)}^{2} \\ \Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \end{array}$

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