So basically the problem is that we have a satellite that is orbiting around this equation:
$3x^2-2xy+7y^2 - 20 = 0$
And it looks like this in graph form: the orbit of the so said planet
At what points can the satellite that is leaving the orbit in a straight line to pass the point $(4,0)$ on the way out into space? I think I remember something with setting up all the equations that must apply to the $x$ value at this point and then solving that equation system. But I can't remember how to do so.
TL;DR Finding at which point our satellite will need to be to hit $(4,0)$.
Sorry for my bad English; it's my third language so I hope I made myself clear.
Thanks in advance!
$$y=m(x-4)$$ is the equation of a generic line passing through $(4,0)$
If we want the intersection points between the line and the ellipse, we must solve the system $$\begin{cases} 3 x^2-2 x y+7 y^2-20=0\\ y=mx-4m\\ \end{cases}$$ substitute the second equation into the first to get $$3x^2-2x(mx-4m)+7(mx-4m)^2-20=0$$ expand and collect $x$ $$\left(7 m^2-2 m+3\right) x^2+\left(8 m-56 m^2\right) x+112 m^2-20=0\tag{1}$$ The line is tangent to the ellipse if this equation has only one root, that means that its discriminant must be zero. $$D=\left(8 m-56 m^2\right)^2-4\left(7 m^2-2 m+3\right)(112 m^2-20)=0$$ $$D=-80 \left(9 m^2+2 m-3\right)=0\to m=\frac{1}{9} \left(-1\pm 2 \sqrt{7}\right)$$ To find the actual points of tangency, substitute the values of $m$ into $(1)$
$$m=\frac{1}{9} \left(-1+ 2 \sqrt{7}\right)\to x=\frac{1}{6} \left(10-\sqrt{7}\right);\;y=\frac{1}{9} \left(-1+ 2 \sqrt{7}\right)\left(\frac{1}{6} \left(10-\sqrt{7}\right)-4\right)=-\frac{\sqrt{7}}{2}$$
$$m=\frac{1}{9} \left(-1- 2 \sqrt{7}\right)\to x=\frac{1}{6} \left(\sqrt{7}+10\right);\;y=\frac{1}{9} \left(-1- 2 \sqrt{7}\right)\left(\frac{1}{6} \left(10+\sqrt{7}\right)-4\right)=\frac{\sqrt{7}}{2}$$ The possible points are $$P_1\left(\frac{1}{6} \left(10-\sqrt{7}\right);\;-\frac{\sqrt{7}}{2}\right);\;P_2\left(\frac{1}{6} \left(10+\sqrt{7}\right);\;\frac{\sqrt{7}}{2}\right)$$