The question:
A curve has equation $4x^2+8x+9y^2-36y+4=0$.
(i) Find $\dfrac{dy}{dx}$.
(ii) Write down the equation(s) of the tangent(s) to the curve that are parallel to
$\qquad$(a) the $x$-axis
$\qquad$(b) the $y$-axis.Answers $\qquad$ (i) $\dfrac{4x+4}{18-9y}\qquad$ (ii) (a) $y=0$ or $y=4\quad$ (b) $x=2$ or $x=-4$.
I got (i) $\frac{dy}{dx}$. But how do I find equation of tangent parallel to x axis? I thought of setting $\frac{dy}{dx} = 0$ but will get $\frac{4x+4}{18-9y}=0$ how do I continue? For parallel to y axis part I set $\frac{dy}{dx} = 1$?