<s>[INST] Common tangents of two ellipses I have two ellipses: $x^2/25+y^2/9=1$ and $x^2/16+y^2/25=1$. I have to find the equations of common tangents.  
I understand that $xx_0/25+yy_0/9=1$ could be the equation of common tangents, ($x_0,y_0$) being the point of intersection. We will have 4 points of intersection thus 4 tangent equations.  
I tried to find the points of intersection by writing 
$x^2/25+y^2/9=x^2/16+y^2/25$ 
$9|x|=16|y|$ 
$x=16/9y$ and $x=-16/9y$ 
By inserting these into one of the ellipses equations, I got 
$y=45/sqrt(481)$ and $y=45-/sqrt(481)$  
I inserted ($x,y$) instead of ($x_0,y_0$) but this doesn't seem to be the right answer. What am I doing wrong? 
 [/INST] There exists a line $y = mx + b$
$\frac {x^2}{25} + \frac {(mx+b)^2}{9} = 1\\
(9+ 25m^2) x^2 + 50mbx +25b^2- 225 = 0$
Since the line is tangent
$x = \frac {-50mb \pm \sqrt {(50mb)^2 - 4(9+25m^2)(25b^2 + 225)}}{2(9+25m^2)}$
Since the line is tangent (and not intersecting) the discriminant is $0$
$(50mb)^2 - 4(9+25m^2)(25b^2 - 225) = 0$
Applying the line to the other equation
$(32mb)^2 - 4(25+16m^2)(16b^2 - 400) = 0$
Which simplifies to 
$b^2 - 25m^2 - 9 = 0\\
b^2 - 16m^2 + 25$
Respectively
$m^2 = \frac {16}{9}\\
b^2 = \frac {481}{9}$
$y = \pm \frac {4}{3} x \pm \frac {\sqrt {481}}{3} $
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