Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $(\pm\sqrt{3},3/2)$.

share|cite|improve this answer

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.

share|cite|improve this answer

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


and by exploiting the identity


we obtain the parametrization


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,


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 $\left(\pm\sqrt 3,\dfrac32\right)$.

share|cite|improve this answer
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 $(\pm\sqrt{3},3/2).$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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