Slope of the given line 3x + 2y + 1 = 0 is
(–3/2). Let us locate the point on the curve at which the tangent is parallel to given line.

Differentiating the curve both sides with respect to x we get, 6x – 8y = 0

⇒ = = [since parallel to 3x + 2y = 1]

also the point (x 1 , y 1 ) lies on, 3x 2 – 4y 2 = 72

⇒ 3 – 4 = 72

⇒ 3 – 4 =

⇒ 3(4) – 4 = [as = –2]

⇒ = 9

⇒ y 1 = ± 3

Required points are (–6, 3) and (6, –3)

Distance of (–6, 3) from the given line,

= =

and distance of (6, –3) from the given line,

= = =

Thus, (–6, 3) is the required point.