# Find the parallels to a line which are tangent to an ellipse

Having the equation of a line, how can I find which of its parallels are tangent to an ellipse of equation $x^2 + 9y^2 = 1$?

If the equation of the line is $y = mx + q$, I know that its parallels have equation $y = mx + k$, but if I put this equation in a system with the equation of the ellipse, I get a final equation with two unknowns.

Am I on the right track?

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Does the equation of the ellipse have a special form? For instance, are the major and minor axes parallel to the $x$- and $y$-axes? Is the ellipse centred at $(0,0)$? –  TonyK May 4 '12 at 18:02
@TonyK I've updated the question with the equation of the ellipse. –  user1301428 May 4 '12 at 18:08
Dear @user1301428: we rarely delete questions which have already been answered. –  Mariano Suárez-Alvarez May 4 '12 at 23:01

We look, for example, at the ellipse with equation $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$$ We would like to find the tangent lines with given slope $m$.

Such a line will have equation of the shape $y=mx+k$. Substitute in the equation of the ellipse. After a little simplification, we get $$(b^2+a^2m^2)x^2+(2a^2mk)x +a^2k^2-a^2b^2=0.$$ We want this equation to have a "double root." That happens iff the discriminant is $0$. It is a nuisance to type the discriminant in this case, but recall that the discriminant of the quadratic polynomial $px^2+qx+r$ is $q^2-4pr$.

We end up with an equation for $k$, actually a very simple equation, since the discriminant condition gives us a linear equation for $k^2$.

Remark: There is a much nicer way, which I will not give the details of. Take the ellipse. Scale distances in the $y$ direction until you get a circle. Note what happens to the desired slope $m$ under the scaling. Now find the tangent lines using circle geometry properties. Scale back.

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Of course, when I calculate the discriminant, I have an equation where $k$ is the only unknown. So I was indeed on the right track :) –  user1301428 May 5 '12 at 8:40
Yes, you were on the right track. The only missing ingredient was the "double root." –  André Nicolas May 5 '12 at 8:44
Sure, I only considered it implied because of the tangency condition. –  user1301428 May 5 '12 at 11:56

You should have two equations for two unknowns. Comparing the $y$'s: $$\frac{x^2}{a^2}+\frac{(mx+k)^2}{b^2}=1$$ Comparing the derivative of $y$: $$\frac{2x}{a^2}+\frac{2ym}{b^2}=0$$

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There should be an easier way which does not involve the use of derivatives. –  user1301428 May 4 '12 at 17:58
@user1301428: Easier I am not sure, derivative is pretty easy, particularly if you use implicit differentiation. But I have posted a classical algebra no calculus solution, at least for ellipses of the standard school type. The idea works for any curve given by $Q(x,y)=0$, where $Q$ is a quadratic in $x$ and $y$. So does differentiation. –  André Nicolas May 4 '12 at 18:28