Need help with a conic tangent question? (Hyperbolics)

Question

I need to find the equation of the tangent to the hyperbola $$\frac{x^2}{6}-\frac{y^2}{8}=1$$ at the point $(3,2)$.

I tried doing it by substituting for $y$ but the algebra is not nice at all and I wanted to see how people here would do it.

Answer 1

If would compute the gradient of $x^2/6-y^2/8$ at the given point. If nonzero (which it is), it is perpendicular to the level set.

Answer 2

you don't need to solve for $y$. you can implicitly differentiate $x^2/6 - y^2/8 = 1$ to get $$\dfrac{xdx}{3} - \dfrac{ydy}{4} = 0$$ substitute $x = 3, y = 2$ to find the tangent has slope $$dx - \frac{dy}{2} = 0$$ turn this into the tangent by replacing $dx, dy$ by $(x-3), (y-2)$ so that the equation of the tangent line at (3,2) is $$(x-3) - \frac{(y-2)}{2}= 0.$$