The tangent line to the curve has slope equal to [tex]\frac{\mathrm dy}{\mathrm dx}[/tex]. Use implicit differentiation to find this derivative.
[tex]9x^2-4y^2=32[/tex]
[tex]\implies18x-16y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{-18x}{-16y}=\dfrac{9x}{8y}[/tex]
The given line in slope-intercept form is
[tex]9x+2y-1=0\implies y=\dfrac{1-9x}2[/tex]
and has slope -9/2. Any line parallel to this one has the same slope. The tangent to the hyperbola has this slope at points (x, y) such that
[tex]-\dfrac92=\dfrac{9x}{8y}\implies x=-4y[/tex]
Find the points on the hyperbola where this condition is met.
[tex]9(-4y)^2-4y^2=140y^2=32\implies y^2=\dfrac{32}{140}\implies y=\pm\sqrt{\dfrac8{35}}\implies x=\pm4\sqrt{\dfrac8{35}}[/tex]
Then use the point-slop formula to build the equations of these tangents:
[tex]y\mp\sqrt{\dfrac8{35}}=-\dfrac92\left(x\pm4\sqrt{\dfrac8{35}}\right)[/tex]
