Locus of points of intersection of some tangents of an ellipse

Question

The quadratic $x^2+4y^2-axy=0$, when $a>4$ represents two pair of lines through origin. Let them cut the ellipse $x^2+4y^2=4$ at A, B and C,D. The tangents at A and B or at C and D would be parallel which intersect at infinity. But, the tangents at A and C; A and B; C and B; B and D meet on some curve(s). Let us find these curves.

Let the point of intersection of two such tangents (e.g., at A and C) be E$(h,k)$, then line joining A and C is chord of contact of the ellipse namely $hx+4ky=4$. If we homogenize this with the equation of ellipse, we get the combined equation OA and OC as $$x^2+4y^2=4(xh+4ky)^2/16 \implies (1-h^2/4)x^2+(4-4k^2)y^2-2hkxy=0$$ Now let us compare it with another given combined equation of OA and OC which is $x^2+4y^2-axy=0$, we get $$\frac{1-h^2/4}{1}=\frac{4-4k^2}{4}=\frac{2hk}{a}\implies h^2=4k^2.$$ So we get the locus of point of intersection of these pair of tangents as $x^2=4y^2.$

The question is: How else this locus could be obtained?

Answer 1

Let $y=mx$ be the lines OA and OC represented by $x^2+4y^2-axy=0$,then $m_1,m_2=1/4$ and let $|a|>4$. Further take $m_1=m, m_2=1/(4m)$.

The points A and C are $$A\left(\frac{2}{\sqrt{1+4m^2}},\frac{m}{\sqrt{1+4m^2}} \right),\quad C\left(\frac{4m}{\sqrt{1+4m^2}}, \frac{1}{\sqrt{1+4m^2}}\right).$$ Then equation of tangents at A and C are: $$x+4my=2\sqrt{1+4m^2},\quad mx+y=\sqrt{1+4m^2},$$ Their point of intersection E is: $$E\left(x=\frac{2\sqrt{1+4m^2}}{1+2m},y=\frac{\sqrt{1+4m^2}}{1+2m}\right)$$ By eliminating $m$. we get the required locus of point of intersection of tangents as $x^2=4y^2.$

Answer 2

It is not difficult to find that the chords of contact have fixed slopes $\pm{1\over2}$ (can give details if needed). That means that their midpoints are all aligned with the origin and with the intersections of the corresponding tangents. But for $a\to+\infty$ the intersection points are $(\pm2,1)$, hence the locus is formed by the lines of equation $y=\pm{1\over2}x$.