If the hyperbola kx^2 - 16y^2 = 144 has an eccentricity of 5 4 and passes through a point P in the first quadrant whose x -coordinate is 5 , then the tangent to the hyperbola at P forms a triangle with the coordinate axes. The value of 15 Area ( OAB) , where O is the origin and A, B are the points of intersection of the tangent with the axes, is

Step-by-step solution

The given equation of the hyperbola is kx^2 - 16y^2 = 144 , which can be rewritten as x^2 144 k - y^2 9 = 1 . Here, a^2 = 144 k and b^2 = 9 . The eccentricity e is given by e^2 = 1 + b^2 a^2 . Substituting the given values, we get ( 5 4 )^2 = 1 + 9 144 k 25 16 = 1 + k 16 k = 9 . So, the equation of the hyperbola is 9x^2 - 16y^2 = 144 . Let P(5, y) be the point on the hyperbola in the first quadrant. Substituting x = 5 into the equation: 9(25) - 16y^2 = 144 16y^2 = 225 - 144 = 81 y^2 = 81 16 . Since P is in the firs