I was solving a few problems from a textbook and I came across this one:
Find the tangent lines to the graph of $x^2+4y^2 = 36$ that go through
the point $P=(12,3)$
I could find the tangency points $$ but my solution was completely messy (I had four equations that I had to go back-and-forth with.
Is there an elegant/simple solution to that?
Tangency points:
$$P_1 = (0,3)$$
$$P_2 = \left( \frac{24}{5} , 3-\frac{24}{5} \right)$$
Best Answer
Let $y = 3 + m(x-12)$ be the equation of a generic line through $P$. The tangency condition is that the quadratic equation $x^2 + 4(3+m(x-12))^2 = 36$ have a double root (or "repeated" root) - that is the algebraic equivalent of tangency. So set the discriminant equal to zero; this will give you the two values of $m$ for which the line is tangent to the ellipse, and then you can solve for the tangency points.