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
$x=16/9y$ 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?

  • $\begingroup$ You should be looking for lines that are simultaneously tangent to both ellipses, not at the tangents at the ellipses’ intersections. $\endgroup$
    – amd
    Commented Jan 31, 2018 at 2:40
  • $\begingroup$ Right, but where do I start? $\endgroup$
    – Lowkey
    Commented Jan 31, 2018 at 2:51
  • $\begingroup$ I have $xx_0/25+yy_0/9=xx_1/16+yy_1/25$, how do I find $(x_0,y_0)$ and $(x_1,y_1)$ ? $\endgroup$
    – Lowkey
    Commented Jan 31, 2018 at 2:55
  • $\begingroup$ Instead of trying to find points on the ellipses, one approach is to find the tangents to each ellipse that pass through an arbitrary exterior point, then solve for the points at which these pairs of lines coincide. Alternatively, if you know about dual conics, then this problem is equivalent to finding the intersections of the duals of the two ellipses. $\endgroup$
    – amd
    Commented Jan 31, 2018 at 3:05

3 Answers 3


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

$b^2 - 25m^2 - 9 = 0\\ b^2 - 16m^2 + 25$


$m^2 = \frac {16}{9}\\ b^2 = \frac {481}{9}$

$y = \pm \frac {4}{3} x \pm \frac {\sqrt {481}}{3} $


This problem is most easily solved using dual conics. Since the origin is interior to both ellipses, you can assume w.l.o.g. that the equations of the tangents are of the form $\lambda x+\mu y=1$. For this to be tangent to the first ellipse, the coefficients must satisfy the dual conic equation $25\lambda^2 + 9\mu^2 = 1.$ You can derive this equation by setting $\lambda = x_0/25$ and $\mu=y_0/9$ and applying the constraint that $(x_0,y_0)$ lies on the ellipse. Similarly, a tangent to the second ellipse must satisfy the dual conic equation $16\lambda^2+25\mu^2=1$. The coefficients of the equations of the common tangents are the solutions to this system of equations—essentially, you compute the intersection of the two dual conics. The solutions are $\lambda = \pm \frac4{\sqrt{481}}$ and $\mu = \pm\frac3{\sqrt{481}}$, so the common tangents are $$\pm4x \pm3y=\sqrt{481}.$$


$y=mx+n$ is a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ iff $n^2=a^2m^2+b^2.$

Now, solve the following system: $$n^2=25m^2+9$$ and $$n^2=16m^2+25.$$


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