We first find the equation of the ellipse. Dividing both sides of the given equation by 144, we get:

(x^2/9) + (y^2/16) = 1

So, the semi-major and semi-minor axes are a=3 and b=4, respectively.

Now, we find the equations of the tangents. Let the point of contact of PQ be (a,b). Then, we have:

(a^2/9) + (b^2/16) = 1 (on the ellipse)
Slope of PQ = (b-4)/(a-3) (tangent to the ellipse is perpendicular to the radius)

Using point-slope form, we get the equation of PQ as:

y-4 = [(b-4)/(a-3)](x-3)

Similarly, for the point of contact of PR, let it be (c,d):

(c^2/9) + (d^2/16) = 1 (on the ellipse)
Slope of PR = (d-4)/(c-3) (tangent to the ellipse is perpendicular to the radius)

Using point-slope form, we get the equation of PR as:

y-4 = [(d-4)/(c-3)](x-3)

Now, we need to find the angle between PQ and PR. Let the angle be θ. Then, we have:

tan(θ) = |[m1-m2]/[1+m1m2]|

where m1 and m2 are the slopes of PQ and PR, respectively.

Substituting the values, we get:

tan(θ) = |[(d-4)/(c-3)] - [(b-4)/(a-3)]| / [1 + (d-4)/(c-3) * (b-4)/(a-3)]

Simplifying, we get:

tan(θ) = |(d-c)/(1+cd) - (b-a)/(1+ab)|

We know that PQ and PR are tangents from the same point P(3,4), so they must be symmetric about the line passing through P and the center of the ellipse (0,0). The equation of this line is y = (4/3)x.

Let the foot of perpendiculars from (a,b) and (c,d) to this line be (p,q) and (r,s), respectively. Then, we have:

q = (4/3)p
s = (4/3)r

Using the distance formula, we get:

p^2 + q^2 = a^2
r^2 + s^2 = c^2

Adding these equations, we get:

p^2 + q^2 + r^2 + s^2 = a^2 + c^2

Substituting the values of q and s, and simplifying, we get:

(p-r)^2 + (16/9)(p+r)^2 = 9

Let x = p-r, y = (4/3)(p+r). Then, we have:

x^2 + y^2 = 9

So, the locus of (p,q) and (r,s) is a circle centered at (0,0) with radius 3.

Now, we use the fact that PQ and PR are symmetric