I am told to find the two tangent lines to the ellipse that pass through the origin, but have been stuck for far too long with my approach, hence am thinking that my approach may be flawed. Here is what I have so far:

If I interpret the ellipse as the level curve of some function $f(x,y)=1$, then I can use the fact that the gradient vector is perpendicular to the ellipse at every point $(a,b)$ on it. Computing the partials, I get that the gradient vector at any point $(a,b)$ is $$\nabla f(a, b) = \left( 2(a-3), \frac{(b-4)}{2}\right),$$ thus, the tangent vector at $(a,b)$ is $$\left(-\frac{(b-4)}{2}, 2(a-3)\right),$$ thus the equation of any tangent line to the ellipse passing through the origin is $$k\left(-\frac{(b-4)}{2}, 2(a-3)\right), k\in \mathbb Z.$$ However, I really don't get how I'm supposed to find... another tangent line? Have I made an oversight in one of the steps of my reasoning? To resolve this, I tried also tried the approach of parametrizing the ellipse into a vector-valued function $$\vec r(t)=(\cos t +3, 2\sin t +4),$$ which can equivalently be interpreted as the orbit of some moving partical. Then, I can differentiate this to get the velocity function of the particle: $$\vec r(t) = (-2\sin t,\cos t).$$ But then, how am I to ensure that the tangent lines pass through the origin?

  • $\begingroup$ $((x-3)^2+\frac14(y-4)^2-1)\cdot ((-3)^2+\frac14 (-4)^2-1)-(-3 x+9+\frac14(-4 y+16)-1)^2= 3 x^2-6 x y+2 y^2=-\frac13 (-3 x + (\sqrt3 + 3)y) (3 x + (\sqrt3 - 3)y)=0$ from Joachimsthal. $\endgroup$ Mar 30 at 10:13
  • $\begingroup$ math.stackexchange.com/questions/440418/… $\endgroup$ Mar 30 at 10:46
  • $\begingroup$ Question: does the question require a vector approach or will implicit differentiation be permitted? $\endgroup$
    – Red Five
    Mar 30 at 11:13
  • $\begingroup$ I think ideally a vector approach, but I may be wrong $\endgroup$
    – Jason Xu
    Mar 30 at 11:33

3 Answers 3


From your first method, you know that a vector of the form $$\vec t = \left(-\frac{1}{2} k(b-4), 2k(a-3) \right)$$ is tangent to the ellipse at the point $(a,b)$. Such a tangent will pass through the origin if there exists a $k \ne 0$ such that $(a,b) = \vec t$. The reason is because if you imagine drawing such a tangent from the origin, it will pass through $(a,b)$ and be parallel to the vector $\vec t$. So for a suitable scaling constant $k$, the vector $(a,b)$ will be equivalent to $\vec t$. Moreover, we require $(a,b)$ to be on the ellipse.

Hence we require solutions of the system $$\begin{align} -\frac{1}{2} k(b-4) &= a, \\ 2k(a-3) &= b, \\ (a-3)^2 + \frac{(b-4)^2}{4} &= 1. \end{align}$$ This gives us $$(a,b,k) \in \left( \frac{4}{13}(9 \pm \sqrt{3}), \; \frac{12}{13}(4 \mp \sqrt{3}), \; \pm 2 \sqrt{3} \right)$$ where the signs must be taken either $(+,-,+)$ or $(-,+,-)$.


I would split the ellipse into $2$ separate functions, then solve for the tangent line(s) of each function.

$T(x)=4+\sqrt{4-4(x-3)^2} \text{ (the top half of the ellipse)}\\ B(x)=4-\sqrt{4-4(x-3)^2} \text{ (the bottom half of the ellipse)}$

Next, we differentiate them:

$\frac{dT}{dx}=\frac{-4(x-3)}{\sqrt{4-4(x-3)^2}}\\ \frac{dB}{dx}=\frac{4(x-3)}{\sqrt{4-4(x-3)^2}}$

These derivatives tell us the slope of their respective halves of the ellipse at any instantaneous point $(a,b)$, and consequently, the slope of the tangent line at that point. Now, in order for the tangent line to pass through the origin, the slope of the tangent line at $(a,b)$ must be equal to the slope of the line joining the points $(0,0)$ and $(a,b)$. We can represent this mathematically, and for now, I'll focus only on $T(x)$ to avoid juggling two things at once:


However, there's just one problem. With this equation, we can't be sure that the point $(a,b)$ really is a point on the ellipse, and not just some random point on the coordinate plane. To fix this, we need to rewrite $b$ in terms of $a$. That way, the point $(a,b)$ would always land on the graph of the ellipse.


Now we can substitute this value of $b$ back into the previous equation and solve for $a$:

$\frac{-4(a-3)}{\sqrt{4-4(a-3)^2}}=\frac{4+\sqrt{4-4(a-3)^2}}{a}\\ -4a(a-3)=4\sqrt{4-4(a-3)^2}+4-4(a-3)^2\\ -4a^2+12a=4\sqrt{4-4(a-3)^2}+4-4a^2+24a-36\\ -3a+8=\sqrt{4-4(a-3)^2}\\ 9a^2-48a+64=4-4(a-3)^2\\ 13a^2-72a+96=0\\ a=\frac{36+4\sqrt{3}}{13},\;\frac{36-4\sqrt{3}}{13}$

Keep in mind that the first solution happens to be extraneous. This is because one of the intermediate calculations was to square both sides of the equation, introducing an additional solution which would not work in the original equation if plugged back in (similar to how $y=\sqrt{x}$ is not the same as $y^2=x$). Therefore, the tangent line of $T(x)$ that passes through the origin intersects $T(x)$ at $x=\frac{36-4\sqrt{3}}{13}$. The equation for the tangent line would be:

$y=\frac{-4(\frac{36-4\sqrt{3}}{13}-3)}{\sqrt{4-4(\frac{36-4\sqrt{3}}{13}-3)^2}}(x-\frac{36-4\sqrt{3}}{13})+(4+\sqrt{4-4(\frac{36-4\sqrt{3}}{13}-3)^2})\; \text{, which $\approx$}\\ y=2.366(x-2.236)+5.291$

Now we just have to repeat this same process for $B(x)$ to see if the bottom half of the ellipse has another possible tangent line that meets the condition.

$\frac{4(a-3)}{\sqrt{4-4(a-3)^2}}=\frac{4-\sqrt{4-4(a-3)^2}}{a}\\ 4a(a-3)=4\sqrt{4-4(a-3)^2}-4+4(a-3)^2\\ 4a^2-12a=4\sqrt{4-4(a-3)^2}-4+4a^2-24a+36\\ 3a-8=\sqrt{4-4(a-3)^2}\\ 9a^2-48a+64=4-4(a-3)^2\\ 13a^2-72a+96=0\\ a=\frac{36+4\sqrt{3}}{13},\;\frac{36-4\sqrt{3}}{13}$

This time, the latter solution is extraneous, for the same reason we saw earlier. Therefore, the tangent line of $B(x)$ that passes through the origin touches $B(x)$ at $x=\frac{36+4\sqrt{3}}{13}$. The equation of this tangent line would be:

$y=\frac{4(\frac{36+4\sqrt{3}}{13}-3)}{\sqrt{4-4(\frac{36+4\sqrt{3}}{13}-3)^2}}(x-\frac{36+4\sqrt{3}}{13})+(4-\sqrt{4-4(\frac{36+4\sqrt{3}}{13}-3)^2})\; \text{, which $\approx$}\\ y=0.634(x-3.302)+2.0935$

So, the tangent lines of the ellipse that also pass through the origin are approximately:

$y=2.366(x-2.236)+5.291\;\; \text{, and}\\ y=0.634(x-3.302)+2.0935$

  • $\begingroup$ By the way, I upvoted this question because it gave me a really nice challenge where I could implement the knowledge I had, even if my answer seems relatively elementary compared to other possible approaches. Thanks for this :) $\endgroup$
    – VV_721
    Apr 2 at 21:56

The tangent lines to the ellipse whose equation is

$ (r - C)^T Q (r - C) = 1$

and passing through point $P$, are given by the quadratic equation

$ \bigg((P - C)^T Q (r - C)-1\bigg)^2 = \bigg((P-C)^T Q (P-C) - 1\bigg) \bigg( (r - C)^T Q ( r - C) - 1\bigg) $

where $ r = [x,y]^T $. In this problem, we have

$ Q = \begin{bmatrix} 1 && 0 \\ 0 && \dfrac{1}{4} \end{bmatrix} $

and $ C = (3, 4) $. And $P = (0,0) $, hence the two tangents (collectively) are given by

$ \bigg( [-3, -4] \begin{bmatrix} 1 && 0 \\ 0 && \dfrac{1}{4} \end{bmatrix} \begin{bmatrix} x - 3 \\ y - 4 \end{bmatrix} - 1 \bigg)^2 = \bigg( (-3)^2 + \dfrac{(-4)^2}{4} - 1\bigg) \bigg( (x - 3)^2 + \dfrac{(y - 4)^2}{4} - 1 \bigg) $

Expanding leads to,

$ \bigg( -3 ( x - 3) - (y - 4) - 1 \bigg)^2 = 12 \bigg( (x - 3)^2 + \dfrac{(y - 4)^2}{4} - 1 \bigg) $

This simplifies to

$ \bigg( -3 x - y + 12 \bigg)^2 = 12 \bigg( x^2 - 6 x + \dfrac{y^2}{4} - 2 y + 12\bigg) $

Expanding gives

$ 9 x^2 + y^2 + 144 + 6 x y - 72 x - 24 y = 12 x^2 - 72 x + 3 y^2 - 24 y + 144 $

And this reduces to

$ 3 x^2 + 2 y^2 - 6 x y = 0 $

This is the equation of the two tangent lines combined. Now, we just have to factor the left hand side:

$ 3 x^2 - 6 x y + 2 y^2 = 0 $

Divide by $x^2$:

$ 2 (y/x)^2 - 6 (y/x) + 3 = 0 $


$ y/x = \dfrac{ 6 \pm \sqrt{36 - 24} } { 4} = \dfrac{3 \pm \sqrt{3} }{2} $

Therefore, the tangent lines are

$ y = \bigg( \dfrac{3 - \sqrt{3}}{2} \bigg) \ x $


$ y = \bigg( \dfrac{3 + \sqrt{3}}{2} \bigg) \ x $


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