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

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up vote 1 down vote accepted

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).

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Procedure to find horizontal tangents:

Let a straight line

$ y = c ... $ (1*)

cut the given ellipse with equation

$ 4x^2+8x+9y^2-36y+4=0$...(2*)

$ 4x^2+8x+9c^2-36c+4=0$...(3*), which is a quadratic equation in x that should have equal roots at point of tangency.

So its determinant must vanish.

$ 64 - 4\,.4\, (9c^2-36c+4) = 0$

Simplify and solve,

$ c = 0,$ and $ c= 4, $ or,

$ y=0 $ and $ y= 4 $ ...(4*)

are the required horizontal tangents.

Similarly proceed setting $ x = c $ for vertical tangents.

Some work can be avoided, or at least reduced by casting the ellipse into standard form:

$ \left(\dfrac{x+1}{3}\right)^2 + \left(\dfrac{y-2}{2}\right)^2 =1 $...(5*)

By adding semi major/minor axes $ (3,2) $to the center of ellipse x,y coordinates $ (-1,2) $ , you know where your tangent points are.

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