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

Question

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:

$$P_1 = (0,3)$$ $$P_2 = \left( \frac{24}{5} , 3-\frac{24}{5} \right)$$

Answer 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.

Answer 2

Just check how tangent of an ellipse derivation by differentiation is done:

$$ \frac{x x_1}{a^2}+ \frac{y y_1} {b^2} =1,$$

so you can just plug in the 4 numbers for $ x_1, y_1, a,b .$

Answer 3

$x^2+4y^2 = 36\\ (\frac{x}{6})^2 + (\frac{y}{3})^2 = 1\\ u = \frac{x}{6}, v = \frac{y}{3}\\ u^2 + v^2 = 1$

$(12,3)$ transfoms to $(2,1)$ in $(u,v)$ space.

$(0,1)$ is one point of tangency such that the tangent goes through $(2,1).$ and if we double the angle between the line from the center to the point of tangency and the line to $(2,1)$ we will find our other point of tangency $(4/5,-3/5)$

And transform back to $(x,y)$ space

Tangent points: $(0,3)$ and $(24/5,-9/5)$

tangents $y = 3$ and $3(y-3) = 2(x-12)$