Show that the set of all points such that the difference of their distances from (4, 0) and (− 4,0) is always equal to 2 represents a hyperbola.
Solution
Let the point be P(x, y)
\[\therefore \left[ \sqrt{\left( x - 4 \right)^2 + \left( y - 0 \right)^2} \right] - \left[ \sqrt{\left( x + 4 \right)^2 + \left( y - 0 \right)^2} \right] = 2\]
\[ \Rightarrow \left[ \sqrt{\left( x - 4 \right)^2 + \left( y - 0 \right)^2} \right]^2 = \left[ 2 + \sqrt{\left( x + 4 \right)^2 + \left( y - 0 \right)^2} \right]^2 \]
\[ \Rightarrow \left( x - 4 \right)^2 + y^2 = 4 + \left( x + 4 \right)^2 + y^2 + 4\sqrt{\left( x + 4 \right)^2 + \left( y - 0 \right)^2}\]
\[ \Rightarrow \left( x - 4 \right)^2 - \left( x + 4 \right)^2 = 4 + 4\sqrt{\left( x + 4 \right)^2 + \left( y - 0 \right)^2}\]
\[\Rightarrow - 16x = 4 + 4\sqrt{\left( x + 4 \right)^2 + \left( y - 0 \right)^2}\]
\[ \Rightarrow - 16x - 4 = 4\sqrt{\left( x + 4 \right)^2 + \left( y - 0 \right)^2}\]
\[ \Rightarrow - 4\left( 4x + 1 \right) = 4\sqrt{\left( x + 4 \right)^2 + \left( y - 0 \right)^2}\]
\[ \Rightarrow - \left( 4x + 1 \right) = \sqrt{\left( x + 4 \right)^2 + \left( y - 0 \right)^2}\]
\[ \Rightarrow 16 x^2 + 8x + 1 = x^2 + 8x + 16 + y^2 \]
\[ \Rightarrow 15 x^2 - y^2 = 15\]
\[ \Rightarrow \frac{x^2}{1} - \frac{y^2}{15} = 1\]
Which is the equation of a hyperbola.
