I have to show that:

If an ellipse and a hyperbola have the same foci, then at each point of intersection, their tangent lines are perpendicular.

So I know that if I prove it for one of the points of intersection then it's valid for all four, but I'm sort of stuck; I don't know what to use.


Let the foci be $F_1$ and $F_2$ and let a point of intersection be $P$. The tangent to the ellipse is the external angle bisector of $\angle F_1PF_2$, and the tangent to the hyperbola is the internal angle bisector.

Proving these statements without calculus is onerous. Here's a sketch of a partial proof for the statement about ellipses, to give the main ideas:


Let $F_1,F_2,P$ be as above. Let $b$ be the external bisector of $\angle F_1PF_2$; we want to show that $b$ is tangent to the ellipse. By the definition you've been given, that means we want to show that $b$ intersects the ellipse only at $P$. Consider first a point $Q$ on the other side of $b$ from the foci; I claim that $Q$ is not on the ellipse. To show this, draw $F_2Q$, meeting $b$ at $R$; also reflect $F_2$ in $b$ to obtain its image $F_2'$. Note that $F_1PF_2'$ are collinear because $b$ is the external angle bisector. Then $$ F_1Q+QF_2 = F_1Q + QR + RF_2 = F_1Q + QR + RF_2' > F_1F_2' = F_1P + PF_2' = F_1P + PF_2 $$ So $Q$ is not on the ellipse. I'm skipping some details here, notably why it has to be $>$ and not $\ge$ in the triangle inequality here. After you fill in that detail, and if you believe that the ellipse is a continuous curve, then we know that the ellipse doesn't cross $b$, so it is tangent to the ellipse. Then give a similar analysis for hyperbolas.

(We have not proved that conic sections have only one tangent at each point; I think I proved that once from your definition, but I only remember that it was unpleasant... and maybe my proof was wrong anyway.)

The nice way to prove these statements is with the notion of gradient from multivariable calculus: the direction of the gradient of the scalar field "distance from $F_i$" is obviously a unit vector pointing away from $F_i$; from this and the definition of an ellipse as a level curve of the sum of two such scalar fields it's clear that the normal to the ellipse bisects the desired angle. (On preview, along the lines of achille hui's answer.)

  • 1
    $\begingroup$ For more proofs using calculus to obtain the tangents analytically, see this question. $\endgroup$ – user21467 Nov 17 '14 at 19:03
  • $\begingroup$ thanks a lot, this is really clear $\endgroup$ – André Nov 17 '14 at 19:04
  • $\begingroup$ No problem. I forgot to mention: the definition you've been given for tangents doesn't handle parabolas; a line parallel to the axis of a parabola intersects it only once but is not a tangent. (This can be fixed by working in projective geometry, but I assume that's not where you're at.) $\endgroup$ – user21467 Nov 17 '14 at 19:13
  • $\begingroup$ one question, at which point are we using that b is the angle bisector? $\endgroup$ – André Nov 20 '14 at 20:18
  • $\begingroup$ Good question! It's when we say that $F_1F_2' = F_1P+PF_2'$, which assumes the three points are collinear. To prove that needs our assumption on $b$. $\endgroup$ – user21467 Nov 20 '14 at 21:20


WLOG we can assume the equations of the ellipse & the hyperbola be $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \frac{x^2}{A^2}-\frac{y^2}{B^2}=1$$ respectively

with foci being $(\pm ae,0);(\pm AE,0)$ where $e,E$ are the eccentricity respectively

We have $ae= AE$

Solve for $x,y$

Hope you know how to find the equation of a tangent of a hyperbola or an ellipse


Let $r_1(p)$ and $r_2(p)$ be the distance of a point $p$ from two fixed foci $f_1$ and $f_2$.

The ellipse / hyperbola passing through $p$ having $f_1$ and $f_2$ as foci are those points $q$ satisfying

$$r_1(q) + r_2(q) = r_1(p) + r_2(p)\quad\text{ and }\quad r_1(q) - r_2(q) = r_1(p) - r_2(p)$$

This implies the normal vectors for the ellipse / hyperbola at point $p$ are in the direction

$$\nabla (r_1 + r_2)\quad\text{ and }\quad\nabla (r_1 - r_2)$$

Notice $$\nabla (r_1 + r_2) \cdot \nabla (r_1 - r_2) = |\nabla r_1 |^2 - |\nabla r_2|^2 = 1 - 1 = 0$$ The normal vectors for the ellipse and hyperbola are perpendicular to each other. As a result, so do their tangent lines.

  • $\begingroup$ what does the triangle mean? $\endgroup$ – André Nov 17 '14 at 18:53
  • $\begingroup$ It means the gradient vector. eg. if you have a function $r = \sqrt{x^2+y^2+z^2}$, then $\nabla r = \left(\frac{\partial r}{\partial x},\frac{\partial r}{\partial y},\frac{\partial r}{\partial z}\right) = \left(\frac{x}{r},\frac{y}{r},\frac{z}{r}\right)$. I come up this answer before I notice you mention you don't want to use calculus in another answer. $\endgroup$ – achille hui Nov 17 '14 at 18:56

You might as well choose the foci to be something convenient, say $(-1,0)$ and $(1,0)$. Then the equation of the ellipse is $\frac {x^2}{a^2}+\frac {y^2}{b^2}=1$ with the requirement (per Wikipedia) that $\sqrt{a^2-b^2}=1$ to make the foci be there so the equation becomes $\frac {x^2}{a^2}+\frac {y^2}{a^2-1}=1$. Now write a similar equation for a hyperbola with the same foci. It should also have one parameter. Solve the two equations to find the intersection point and differentiate to find the slopes of the tangents.

  • $\begingroup$ I'm not supposed to use calculus $\endgroup$ – André Nov 17 '14 at 18:06
  • $\begingroup$ I didn't see that in the question. You need Steven Taschuk's solution, then, but I don't know how to justify the claims. They do seem reasonable. $\endgroup$ – Ross Millikan Nov 17 '14 at 18:11
  • $\begingroup$ Actually, I don't know how to justify my claims without calculus either. @André, what are you allowed to use about tangents to conic sections? $\endgroup$ – user21467 Nov 17 '14 at 18:15
  • $\begingroup$ Well, they give me this definition: "The angle a in P between two curves A and B is the angle formed by their tangents in P. A tangent is a line that touches the curve in a single point." $\endgroup$ – André Nov 17 '14 at 18:23
  • $\begingroup$ OK, I'll expand my answer using that. $\endgroup$ – user21467 Nov 17 '14 at 18:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.