I am told to find the two tangent lines to the ellipse that pass through the origin, but have been stuck for far too long with my approach, hence am thinking that my approach may be flawed. Here is what I have so far:
If I interpret the ellipse as the level curve of some function $f(x,y)=1$, then I can use the fact that the gradient vector is perpendicular to the ellipse at every point $(a,b)$ on it. Computing the partials, I get that the gradient vector at any point $(a,b)$ is
$$\nabla f(a, b) = \left( 2(a-3), \frac{(b-4)}{2}\right),$$ thus, the tangent vector at $(a,b)$ is
$$\left(-\frac{(b-4)}{2}, 2(a-3)\right),$$ thus the equation of any tangent line to the ellipse passing through the origin is $$k\left(-\frac{(b-4)}{2}, 2(a-3)\right), k\in \mathbb Z.$$ However, I really don't get how I'm supposed to find… another tangent line? Have I made an oversight in one of the steps of my reasoning? To resolve this, I tried also tried the approach of parametrizing the ellipse into a vector-valued function $$\vec r(t)=(\cos t +3, 2\sin t +4),$$ which can equivalently be interpreted as the orbit of some moving partical. Then, I can differentiate this to get the velocity function of the particle: $$\vec r(t) = (-2\sin t,\cos t).$$ But then, how am I to ensure that the tangent lines pass through the origin?
Best Answer
From your first method, you know that a vector of the form $$\vec t = \left(-\frac{1}{2} k(b-4), 2k(a-3) \right)$$ is tangent to the ellipse at the point $(a,b)$. Such a tangent will pass through the origin if there exists a $k \ne 0$ such that $(a,b) = \vec t$. The reason is because if you imagine drawing such a tangent from the origin, it will pass through $(a,b)$ and be parallel to the vector $\vec t$. So for a suitable scaling constant $k$, the vector $(a,b)$ will be equivalent to $\vec t$. Moreover, we require $(a,b)$ to be on the ellipse.
Hence we require solutions of the system $$\begin{align} -\frac{1}{2} k(b-4) &= a, \\ 2k(a-3) &= b, \\ (a-3)^2 + \frac{(b-4)^2}{4} &= 1. \end{align}$$ This gives us $$(a,b,k) \in \left( \frac{4}{13}(9 \pm \sqrt{3}), \; \frac{12}{13}(4 \mp \sqrt{3}), \; \pm 2 \sqrt{3} \right)$$ where the signs must be taken either $(+,-,+)$ or $(-,+,-)$.