Question

The locus of a point whose difference of distance from points (3, 0) and (–3,0) is 4, is.

• A. $\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1$
• B. $\frac{x^{2}}{5} - \frac{y^{2}}{4} = 1$
• C. $\frac{x^{2}}{2} - \frac{y^{2}}{3} = 1$
• D. $\frac{x^{2}}{3} - \frac{y^{2}}{2} = 1$

Answer 1

Answer: (A)

$\frac{x^{2}}{4} - \frac{y^{2}}{5} = 1$

Answer 2

Let the point be

Given that $P A - P B = 4$

$\sqrt { ( h - 3 ) ^ { 2 } + k ^ { 2 } } - \sqrt { ( h + 3 ) ^ { 2 } + k ^ { 2 } } = 4$

$\Rightarrow \sqrt { ( h - 3 ) ^ { 2 } + k ^ { 2 } } = 4 + \sqrt { ( h + 3 ) ^ { 2 } + k ^ { 2 } }$

Squaring both sides, we get

$( h - 3 ) ^ { 2 } + k ^ { 2 } = 16 + ( h + 3 ) ^ { 2 } + k ^ { 2 } + 8 \sqrt { ( h + 3 ) ^ { 2 } + k ^ { 2 } }$

$\Rightarrow h ^ { 2 } + 9 - 6 h + k ^ { 2 } = 16 + h ^ { 2 } + 9 + 6 h + k ^ { 2 }$ $+ 8 \sqrt { ( h + 3 ) ^ { 2 } + k ^ { 2 } }$

$\Rightarrow - 6 h = 16 + 6 h + 8 \sqrt { ( h + 3 ) ^ { 2 } + k ^ { 2 } }$ $\Rightarrow - 8 \sqrt { ( h + 3 ) ^ { 2 } + k ^ { 2 } } = 12 h + 16$

Again, squaring both sides, we get

$64 \left( h ^ { 2 } + 9 + 6 h + k ^ { 2 } \right) = 144 h ^ { 2 } + 256 + 2.16 .12 h$ $\Rightarrow 4 \left( h ^ { 2 } + 9 + 6 h + k ^ { 2 } \right) = 9 h ^ { 2 } + 16 + 24 h$ $\Rightarrow 4 h ^ { 2 } + 36 + 24 h + 4 k ^ { 2 } = 9 h ^ { 2 } + 16 + 24 h$

$\Rightarrow 5 h ^ { 2 } - 4 k ^ { 2 } = 20 \Rightarrow \frac { h ^ { 2 } } { 4 } - \frac { k ^ { 2 } } { 5 } = 1$

Hence, the locus of point P is $\frac { x ^ { 2 } } { 4 } - \frac { y ^ { 2 } } { 5 } = 1$.