# How to find points of tangency on an ellipse?

The problem I have to solve is: If tangent lines to ellipse $9x^2+4y^2=36$ intersect the y-axis at point $(0,6)$, find the points of tangency.

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You are missing some $y^2$ somewhere in your expression. Can you take a look at it again? – smanoos Oct 16 '11 at 7:15
None of the (multivariable-calculus), (elliptic-curves) or (elliptic-functions) tags are representative of this question. – anon Oct 16 '11 at 9:13

Using implicit differentiation, you can find the slope of a line tangent to the ellipse at a point $(a,b)$ on the ellipse. Taking the derivative, $$18x+8y\frac{dy}{dx}=0\implies \frac{dy}{dx}=\frac{-9x}{4y}.$$ So the slope of a tangent line at a point $(a,b)$ is $\frac{-9a}{4b}$, so the equation of such is line is $$y-b=\frac{-9a}{4b}(x-a)$$ which is equivalent to $$9ax+4by=9a^2+4b^2=36.$$ Since this line must also pass through $(0,6)$, plugging in you find $24b=36$, or $b=3/2$. Substituting back into the original equation yields $a=\pm\sqrt{3}$, so the two points of tangency are $(\pm\sqrt{3},3/2)$.

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A standard way to solve the problem is to consider the generic line passing through $(0,6)$ which has equation $$y-6=mx,$$ then make a substitution in the ellipse equation and impose that the resulting one variable quadratic equation has a double root. This will give you the values $m$ of the tangent lines.

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Yet another method: converting the equation of the ellipse into the form

$$\frac{x^2}{4}+\frac{y^2}{9}=1$$

and by exploiting the identity

$$\left(\frac{1-t^2}{1+t^2}\right)^2+\left(\frac{2t}{1+t^2}\right)^2=1$$

we obtain the parametrization

\begin{align*}x&=2\frac{1-t^2}{1+t^2}\\y&=\frac{6t}{1+t^2}\end{align*}

From this parametrization, we derive the equation of the tangent line:

$$y=\frac{3(t^2-1)}{4t} \left(x-2\frac{1-t^2}{1+t^2}\right)+\frac{6t}{1+t^2}$$

or, simplified,

$$y=\frac{3(t^2-1)}{4t}x+\frac{3\left(t^2+1\right)}{2t}$$

We ask that the $y$-intercept of the tangent line be $6$; equating the constant term of the linear equation to $6$ and rearranging yields

$$3 t^2-12 t+3=0$$

which has the roots $t=2\pm\sqrt{3}$; substituting these values of $t$ into the original parametric equations yields the tangency points $\left(\pm\sqrt 3,\dfrac32\right)$.

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Does that identity (that is being exploited) have a name? – The Chaz 2.0 Oct 25 '11 at 1:08
I actually cheated a bit by implicitly using the Weierstrass substitution here, Chaz. :) – J. M. Oct 25 '11 at 1:10
I was gonna say... if my students tried to exploit such an identity, well... they might need to teach me first! – The Chaz 2.0 Oct 25 '11 at 1:12
I presented it anyway since the Weierstrass substitution is so useful for solving algebraic problems involving trigonometrics. Even when, no, especially when you're using Gröbner bases. – J. M. Oct 25 '11 at 1:15
I see. Cheers!. – The Chaz 2.0 Oct 25 '11 at 1:22

We use a linear transformation. Go from $(3x)^2+(2y)^2=6^2$ to $u^2+v^2=1$ by using a change of coordinates with $u=\frac{1}{2}x$ and $v=\frac{1}{3}y$. Tangencies and intersections are preserved, so there are two lines tangent to the unit circle intersecting the point $(0,2)$ in the $uv$-plane; they will be symmetric across the $v$-axis so finding one is enough. Make a right triangle with one vertex at the origin $O$, one at the point of tangency $T$ (say on the right side), and one at $P=(0,2)$, with $\phi=\angle POT$; then trigonometry dictates $\cos\phi=1/2\implies\phi=\pi/3$, hence the points in the $uv$-plane are $$(\cos(\pi/2-\phi),\sin(\pi/2-\phi))=\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right),\text{ and}$$ $$(-\cos(\pi/2-\phi),\sin(\pi/2-\phi))=\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right).$$ Transform these points back into the $xy$-plane using $x=2u,y=3v$ and obtain $(\pm\sqrt{3},3/2).$

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