Tangent of Ellipse Given the equation of an ellipse $x^2+4y^2=36$, I am asked to find the equation of the 2 tangent lines that intersect at the point $(12,3)$. I am not sure of what to do after doing the implicit differentiation, I tried forming an equation for the gradient, $(y_1-3)/(x_1-12)$ then equate it to $\mathrm{d}y/\mathrm{d}x$. I'm not sure if I'm approaching it the right way but I can't find an answer.
 A: If $ax^2+by^2=c$ then
$2ax dx+2by dy=0$ so
$y'=dy/dx=-ax/(by)$.
Therefore the equation of the tangent at $(u,v)$ is
$(y-u)/(x-v)=-au/(bv)$.
If this line passes through $(x,y)$,
this gives one equation for $(u,v)$.
The second equation is given by
$(u,v)$ on the ellipse, so
$au^2+bv^2=c$.
Solving these gives $u$ and $v$ in terms of
$x, y, a, b,$ and $c$.
A: Since you are interested in just the equations of the two tangents, this approach can sometimes prove useful. Consider a line passing through $(12, 3)$ with a slope $m$. It will have the equation $y = mx + (3 - 12m)$. If this line intersects the ellipse at a point $(u,v)$, then we will have the relation
\begin{align*}
u^2 + 4(mu + (3 - 12m))^2 & = 36 \\
\implies (1 + 4m^2)u^2 + 8m(3 - 12m)u + 4(3 - 12m)^2 - 36 & = 0.
\end{align*}
Moreover, since the line is a tangent, the above quadratic in $u$ should have two repeated roots. Therefore, the discriminant should be zero.  On imposing this condition, you will get a quadratic in $m$ whose two solutions would be the slopes of the two required tangents.
A: [This problem seems to be straight-outta-Stewart:  I recall writing this up for students in the murky past...  I also found in going through problems with the "implicit-differentiation" tag, that this is a much-duplicated question -- see my comment to the OP.]
It would have been helpful if you had written out your implicit differentiation result, but you presumably found something like
$$ \frac{d}{dx} \ [ \ x^2 \ + \ 4y^2 \ ] \ \ = \ \ \frac{d}{dx} \ [ \ 36 \ ] \ \ \Rightarrow \ \ 2x \ + \ 8y· \frac{dy}{dx} \ \ = \ \ 0 \ \ \Rightarrow \ \ \ \frac{dy}{dx} \ \ = \ \ -\frac{x}{4y} \ \ . $$
For a tangent point $ \ (x_1 \ , \ y_1 ) \ $ on the ellipse, you are correct that you want the slope of the line passing through that point and $ \ (12 \ , \ 3) \ $ to be equal to this implicit derivative, so you are right to equate $ \ \large{\frac{y_1 \ - \ 3}{x_1 \ - \ 12} \ = \ -\frac{x_1}{4·y_1}} \ \ . $  We can "cross-multiply" these ratios since we know that the tangent point cannot have $ \ x_1 \ = \ 12 \ \ , \ $ nor can $ \ y_1 \ = \ 0 \ \ . \ $ [The tangent point won't be "directly above or below" the external point $ \ (12 \ , \ 3) \ $ and it can't be on the major axis of the ellipse, since a tangent line at an endpoint is vertical.]
We now have $ \ 4·y_1^2 - 12y_1 \ = \ -x_1^2 + 12·x_1 \ \ . $  Some of these terms seem familiar:  if we re-arrange them and use the ellipse equation, we have
$$ \ x_1^2 \ + \ 4·y_1^2   \ = \  12·x_1 + 12·y_1 \ \ \Rightarrow \ \ 36  \ = \  12·x_1 + 12·y_1 \ \ \Rightarrow \ \ x_1 +  y_1 \ \ = \ \ 3 \ \  . $$
This is the straight line marty cohen alludes to in his answer.  The intersection(s) of this line with the ellipse mark the tangent points we are seeking.  If we insert this into the ellipse equation, we may solve for either coordinate; for instance,
$$ x_1 \ \ = \ \ 3 \ - \ y_1 \ \ \rightarrow \ \ (3 \ - \ y_1)^2 \ + \ 4·y_1^2 \ \ = \ \ 36 \ \ \Rightarrow \ \ (9 \ - \ 6·y_1 \ + \  y_1^2) \ + \ 4·y_1^2 \ \ = \ \ 36 $$
$$ \Rightarrow \ \ 5·y_1^2 \ - \ 6·y_1 \ - \ 27 \ \ = \ \ (5·y_1 \ + \ 9) \ · \ (y_1 \  - \ 3) \ \ = \ \ 0 \ \ . $$
Returning to the intersections equation, $ \ x_1 \ = \ 3 \ - \ y_1 \ \ , $ we obtain the two tangent points $ \ (0 \ , \ 3) \ $ and $ \ \left(\frac{24}{5} \ , \ -\frac95 \right) \ \ . $
The problem asks for the equations of the tangent lines, for which we can write (as your slope ratio suggests), $ \ y - 3 \ = \ \large{\frac{dy}{dx}|_{(x_1 \ , \ y_1)}} · \normalsize{ (x - 12)}   \ \ .  $  We thus obtain
tangent point $ \ (0 \ , \ 3) \ \ : \ \ y - 3 \ = \ \large{-\frac{0}{4 \ · \ 3}} · \normalsize{(x - 12)} \ \ \ $  or $ \ \ \ y \ = \ 3 \ \ ; $
tangent point $ \ \left(\frac{24}{5} \ , \ -\frac95 \right) \ \ : \ \ y - 3 \ = \ \large{-\frac{24/5}{4 \ · \ (-9/5)}} · \normalsize{(x - 12)} \ = \  \frac23 ·   (x - 12)  \ \ \ $
or $ \ \ \ y \ = \ \frac23x \ - \ 5 \ \ . $
The first tangent point is the upper endpoint of the minor axis of the ellipse.  Since $ \ (12 \ , \ 3 ) \ $ has the same $ \ y-$coordinate, it makes sense that this tangent line is "horizontal".
