Find the tangent lines to the graph of $x^2+4y^2 = 36$ that go through the point $P=(12,3)$

Solution 1:

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.

Solution 2:

Scale the ellipse by a factor of $2$ along the $y$-axis so that it becomes a circle: we want to find tangent lines to $x^2+u^2=36$ that pass through $P' = (12,6)$. Let $(x_1,u_1)$ be a point of tangency on such a line. Then the tangent is perpendicular to the line from the origin to $(x_1,u_1)$, so $$\frac{6-u_1}{12-x_1} = -\frac{x_1}{u_1}$$ Simplifying, and using the fact that $x_1^2+u_1^2=36$, we get $2x_1+u_1=6$. Substitute this into the equation for the circle to get $$x_1^2+(6-2x_1)^2=36$$ which reduces to $x_1(5x_1-24)=0$.