How to find equation of a tangent How to find equation of a tangent on a $4x^2+9y^2-24x+18y+9=0$ in $T(6,-1)$?
The solution is $x=6$, but I always get: $y = \frac{-4}{15}x + \frac {3}{5} $(?!)
Alternate form: 

This is the ellipse we get: (if that helps) 

 A: The derivative of $4x^2+9y^2-24x+18y+9=0$ is:
$$8x+18yy'-24+18y'=0$$
or
$$4x+9yy'-12+9y'=0$$
Then,
$$ 9yy'+9y'=12-4x $$
$$9y'(y+1)=12-4x$$
Therefore, $y'=\frac{12-4x}{9(y+1)}$
Now, substitute $x=6, \ y=-1$  to find $y'$(which, as you know, is the slope a point $T$).
At this point you have everything needed finding the equation of a tangent line.
A: HINT:
We don't need to bother whether the point$(6,1)$ lies outside, on or inside the given ellipse.
Equation of any line passing through  $(6,1)$ can be written as $$\frac{y-1}{x-6}=m\ \ \ \  (1)$$ where $m$ is the gradient/slope
Find the intersection of $(1)$ with the  given curve by replacing $y$ with $mx-6m-1$ to form a Quadratic equation in $x$
Now for tangency the two intersection must coincide $\implies$ the discriminant of the Quadratic equation must be equal to zero
The nature of the two values of $m$  will actually determine the number of tangents 
Observe:
If the point lies inside the ellipse, both the values $m$ will complex implying no real tangent
If the point lies on the ellipse, both the values $m$ will be same(real) implying exactly one real tangent
If the point lies outside the ellipse, both the values $m$ will be real & distinct implying two real real tangents
