# Finding Equation of Tangent parallel to x & y axis

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 $\frac{dy}{dx} = 1$?

-

You were correct - by setting $\displaystyle \frac{dy}{dx}= 0$ our find information about which points have that property of having tangent parallel to the $x$-axis. You found that $\displaystyle \frac{4x+4}{18-9y} = 0$ which is only true if $x=-1.$ Plug this into the equation of the curve to find the $y$ values of points on the curve with $x=-1.$
Once you know the coordinates of the points with tangent parallel to the $x$-axis, you automatically know the equation of the tangent. Draw a picture if you don't see how.
For (ii) (b), loosely speaking, the gradient must be infinite, so you must set $\displaystyle \frac{dy}{dx}= \pm \infty$ and go through a similar process as in part a).