Question

Let F_1 = (10,2) and F_ 2= (-16,2). Then the set of points P such that |PF_1 - PF_2| = 24 form a hyperbola. The equation of this hyperbola can be written as ((x-h)^2)/a^2-((y-k)^2)/b^2=1 Find h + k + a + b.

Answer 1

The hyperbola is centered at (-3, 2) with a transverse axis length of 24 and a conjugate axis length of 10. The sum of h, k, a, and b for the hyperbola's equation is 16.

Step-by-step explanation:

The given points F_1 = (10,2) and F_2 = (-16,2) are the foci of the hyperbola, and the condition |PF_1 - PF_2| = 24 describes the set of all points P that form this hyperbola. For a hyperbola centered at (h, k) with a horizontal transverse axis, the foci are at (h ± c, k), where 2c is the distance between the foci. Therefore, the midpoint of F_1 and F_2 will be the center (h, k) of the hyperbola.

We can find the center of the hyperbola by taking the average of the x-coordinates and the y-coordinates of F_1 and F_2, giving us h = (10 - 16)/2 = -3 and k = (2 + 2)/2 = 2. The distance between the foci is 2c, so c = |10 - (-16)|/2 = 13. Using the relationship c² = a² + b² and given that |PF_1 - PF_2| = 2a = 24, we find a = 12 and thus c² = 169 = a² + b², which leads to b² = 169 - 144 = 25 and hence b = 5. The sum h + k + a + b = -3 + 2 + 12 + 5 = 16.