Question

Two points A and B have coordinates (1, 0 )and( -1, 0 )respectively and Q is a point which satisfies the relation |AQ-BQ| = 1 then find the locus of Q​

Answer 1

Answer:

$12x^2-4y^2=3$

Step-by-step explanation:

Let the coordinates of Q be (x, y)

Given |AQ - BQ| = 1

$AQ=\sqrt{(x-1)^2+(y-0)^2} =\sqrt{(x-1)^2+y^2}$

$BQ=\sqrt{(x+1)^2+(y-0)^2} =\sqrt{(x+1)^2+y^2}$

|AQ - BQ| = 1

⇒ AQ - BQ = ±1

$\implies \sqrt{(x-1)^2+y^2}-\sqrt{(x+1)^2+y^2}=\pm1$

$\implies \sqrt{(x-1)^2+y^2}=\pm1+\sqrt{(x+1)^2+y^2}$

squaring both sides

$(x-1)^2+y^2=1\pm2\sqrt{(x+1)^2+y^2}+(x+1)^2+y^2$

$\implies x^2-2x+1+y^2=1\pm2\sqrt{(x+1)^2+y^2}+x^2+2x+1+y^2$

$\implies -4x-1=\pm2\sqrt{(x+1)^2+y^2}$

Squaring both sides again

$16x^2+8x+1=4[(x+1)^2+y^2]$

$\implies 16x^2+8x+1=4[x^2+2x+1+y^2]$

$\implies 16x^2+8x+1=4x^2+8x+4+4y^2$

$\implies 12x^2-4y^2=3$

Which is the locus of Q

The locus of Q is hyperbola

Hope this answer is helpful.