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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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    Dear @user1301428: we rarely delete questions which have already been answered.2012-05-04

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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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    Sure, I only considered it implied because of the tangency condition.2012-05-05
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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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    @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.2012-05-04