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.
How to find points of tangency on an ellipse?
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0You are missing some $y^2$ somewhere in your expression. Can you take a look at it again? – 2011-10-16
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1None of the ([multivariable-calculus](http://en.wikipedia.org/wiki/Multivariable_calculus)), ([elliptic-curves](http://en.wikipedia.org/wiki/Elliptic_curve)) or ([elliptic-functions](http://en.wikipedia.org/wiki/Elliptic_function)) tags are representative of this question. – 2011-10-16
4 Answers
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 $(\pm\sqrt{3},3/2)$.
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.
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 $(\pm\sqrt{3},3/2).$
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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0Does that identity (that is being exploited) have a name? – 2011-10-25
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1I actually cheated a bit by implicitly using the Weierstrass substitution here, Chaz. :) – 2011-10-25
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0I was gonna say... if my students tried to exploit such an identity, well... they might need to teach me first! – 2011-10-25
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0I 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. – 2011-10-25
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0I see. Cheers!. – 2011-10-25