a point moves so that the product of it's directed distances from the lines 3x-4y+1=0 and 3x+4y-7=0 is 144/25. find the equation of it's locus, sketch the curve

Answer 1

The equation of its locus will be the equation of a hyperbola with the two lines as asymptotes. There are two possible hyperbola the equation of the locus can be: a horizontal hyperbola and a vertical hyperbola. We'll look at the vertical hyperbola first. Note that the hyperbola will be centered at (1,1) since this is where our asymptotes intersect. Plug into the equation: b²(y-1)² - a²(x-1)² = pc², where p is the product of the directed distances. The asymptotes have slopes of ±3/4 so a = 3 and b = 4. a²+ b² = c² so c = 5. Next just plug in a, b, c, and p.

Next, we'll look at the horizontal hyperbola. We use a similar equation.

b²(x-1)² - a²(y-1)² = pc².

b in this case is 3 and a = 4. That's because the slope of the asymptotes for a horizontal hyperbola is ± b/a.