# Common point between ellipse and tangent passing through external point

Given an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and a point $(u, v)$ not on the ellipse, I want to find two points that lie on the ellipse and on the two tangents of the ellipse passing through $(u, v)$.

Attempted solution:

Define $f(x, y) = \frac{x^2}{a^2} + \frac{y^2}{b^2} - 1$. A point $(x, y)$ is on the ellipse if $f(x, y) = 0$.

A normal of the ellipse at $(x, y)$ is $\nabla f(x, y) = 2\left(\frac{x}{a^2}, \frac{y}{b^2}\right)$.

The point $(x, y)$ I am looking for has to satisfy two things: (i) it has to lie on the ellipse and (ii) the normal at $(x, y)$ has to be orthogonal to the line connecting $(x, y)$ and $(u, v)$:

(i): $\frac{x^2}{a^2} + \frac{y^2}{b^2} - 1 = 0$

(ii): $\left(\frac{x}{a^2}, \frac{y}{b^2}\right) \cdot (x-u, y-v) = 0$.

Rewriting (ii), we get:

(ii): $\frac{x^2}{a^2} - \frac{ux}{a^2} + \frac{y^2}{b^2} - \frac{vy}{b^2} = 0$.

How to solve (i) and (ii) for $(x, y)$? From (i), we get $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, which we can plug into (ii) to get

(iii) $\frac{ux}{a^2} + \frac{vy}{b^2} = 1$.

But this is an equation of a whole line and not for two points. How to get $(x, y)$ from here? I could solve (iii) for $y$ and then plug the result in (i) to end up with one equation for $x$. But I used up (i) to get to (iii). I cannot go back to (i) from (iii), can I?

• Yes, you can... Commented Jul 27, 2018 at 21:39

• $$(u,v)$$ is known as the pole of the polar (chord), namely

$$\frac{ux}{a^2}+\frac{vy}{b^2}=1$$

• Substitute $$y=\dfrac{b^2}{v}\left( 1-\dfrac{ux}{a^2} \right)$$ into the ellipse:

$$\frac{x^2}{a^2}+\frac{b^2}{v^2}\left( 1-\frac{ux}{a^2} \right)^2=1$$

$$\frac{v^2x^2}{b^2}+a^2\left( 1-\frac{ux}{a^2} \right)^2=\frac{a^2v^2}{b^2}$$

$$\left( \frac{u^2}{a^2}+\frac{v^2}{b^2} \right)x^2- 2ux+a^2\left( 1-\frac{v^2}{b^2} \right)=0$$

• Sum and product of roots:

$$x_1+x_2=\frac{2u}{\dfrac{u^2}{a^2}+\dfrac{v^2}{b^2}}$$

$$x_1 x_2=\frac{a^2\left( 1-\dfrac{v^2}{b^2} \right)}{\dfrac{u^2}{a^2}+\dfrac{v^2}{b^2}}$$

• Points of contact:

$$\left( \frac{u \pm \dfrac{av}{b} \sqrt{\dfrac{u^2}{a^2}+\dfrac{v^2}{b^2}-1}} {\dfrac{u^2}{a^2}+\dfrac{v^2}{b^2}} \, , \, \frac{v \mp \dfrac{bu}{a} \sqrt{\dfrac{u^2}{a^2}+\dfrac{v^2}{b^2}-1}} {\dfrac{u^2}{a^2}+\dfrac{v^2}{b^2}} \, \right)$$

Standard results are summarized in the diagram below where $$A(x',y')$$ is the pole, $$B(x_1,y_1)$$ and $$C(x_2,y_2)$$ are the points of contact.

• ChungTak,1) How did you comute polar length and midpoint of polar(Chord BC)? Commented Dec 14, 2019 at 8:32
• Mid-point $$\left( \frac{x_1+x_2}{2},\frac{y_1+y_2}{2} \right)$$ Length $$\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=\sqrt{(x_1+x_2)^2-4x_1x_2+(y_1+y_2)^2-4y_1 y_2}$$ Sum-and-product of roots is a powerful tool. Commented Dec 14, 2019 at 8:35
• Could you tell me the books you learned this from.. please? Commented Apr 28, 2022 at 7:57
• @Buraian Try to read some old classics from internet archive. Commented Apr 29, 2022 at 2:16

A simple method:

Determine first the slopes of the tangents to the ellipse though the point $(u,v)$, if any.

Consider the pencil of lines throught $(u,v)$. A line of the pencil has equation $$y-v=t(x-u)\iff y=t(x-u)+v \qquad(t\in\bf R).$$ Replace $y$ in the equation of the ellipse, which you ma rewrite as $$b^2x^2+a^2y^2=a^2b^2.$$ This yields the equation $$b^2x^2+a^2\bigl(t(x-u)+v\bigl)^2=a^2b^2.$$ This is a quadratic equation in $x$, with parameter $t$. Its roots are the abscissæ of the intersection points of the line and the ellipse.

Now, the line is tangent to the ellipse if and only if this equation has a double root, i.e. if and only if its discriminant (depending on the parameter) $\;\Delta(t)=0$. Furthermore, the abcissa of the point of contact is the double root, $-b/2a$ with the standard notation for quadratic equations.

• A very efficient way IMHO. Commented Jul 27, 2018 at 22:09
• If $u,v$ are outside the ellipse, that should be a sufficient condition for the roots to exist. Commented Jul 27, 2018 at 22:30