Q.235. To find the equations of the hyperbola with given conditions

Question 235.
A line y=2x+5 intersects a hyperbola only at one point (-2,1). The equation of one of its asymptotes is 3x+2y+1=0. If the hyperbola passes through the point (-1,0), find the equations of all possible hyperbolas.

Answer 235.
There are two possibilities and the equation of the hyperbola is found in each case separately.
 
(1) The line y = 2x + 5 intersects the hyperbola at a single point (-2, 1).
 
As y = 2x + 5 is a line intersecting the hyperbola at one point only, the other asymptote must be parallel to it.
=> the equation of the hyperbola should be of the form
(3x + 2y + 1) (2x - y + k) = k'

As (-2, 1) and (-1, 0) pass through it,
[3*(-2) + 2*(1) + 1] * [2*(-2) - (1) + k] = k' and
[3*(-1) + 2*(0) + 1] * [2*(-1) - (0) + k] = k'
=> k' + 3k = 15 and k' + 2k = 4

Solving, k = 11 and k' = - 18
=> the equation of the required hyperbola is
(3x + 2y +1) (2x - y + 11) = -18
=> 6x^2 + xy - 2y^2 + 35x + 21y + 29 = 0.

(2) The line y = 2x + 5 is a tangent to the hyperbola at the point (-2, 1).

Suppose y = 2x + 5 is a tangent to the hyperbola at (-2, 1).
Let ax + by + c = 0 be the other asymptote
=> the equation of the hyperbola is
(3x + 2y + 1)(ax + by + c) = 1

Points (-2, 1) and (-1, 0) lie on it
=> 6a - 3b - 3c = 1 ... (1)
and 2a - 2c = 1 ... (2)

Differentiating the eqn. of the hyperbola w.r.t. x,
6ax + 4byy' + (a+3c) + (b+2c)y' + (2a+3b)(y+xy') = 0
As the line y = 2x + 5 touches it at (-2, 1) and its slope is y' = 2
=> -12a + 8b + a + 3c + 2b + 4c - 6a - 9b = 0
=> 17a - b - 7c = 0 ... (3)

Solving eqns. (1), (2) and (3) [I used Wolfram Alpha]
a = -10/27, b = -11/54 and c = - 47/54
=> eqn. of the hyperbola is
(3x + 2y + 1)(20x + 11y + 47) = - 54
=> 60x^2 + 22y^2 + 73xy + 161x + 105y + 101 = 0.

Link to YA!