Question

Determine the equation of the tangent of the hyperbola that is parallel to the given line:

Answer 1

The tangent line to the curve has slope equal to $\frac{\mathrm dy}{\mathrm dx}$. Use implicit differentiation to find this derivative.

$9x^2-4y^2=32$

$\implies18x-16y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{-18x}{-16y}=\dfrac{9x}{8y}$

The given line in slope-intercept form is

$9x+2y-1=0\implies y=\dfrac{1-9x}2$

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

$-\dfrac92=\dfrac{9x}{8y}\implies x=-4y$

Find the points on the hyperbola where this condition is met.

$9(-4y)^2-4y^2=140y^2=32\implies y^2=\dfrac{32}{140}\implies y=\pm\sqrt{\dfrac8{35}}\implies x=\pm4\sqrt{\dfrac8{35}}$

Then use the point-slop formula to build the equations of these tangents:

$y\mp\sqrt{\dfrac8{35}}=-\dfrac92\left(x\pm4\sqrt{\dfrac8{35}}\right)$