I'm a beginner in coordinate geometry familiar only with the basics of it. I recently stumbled upon a problem which took many attempts but all in vain. The problem goes as follows:-
If a triangle is inscribed in an ellipse (E$_1$) {major axis $a$ , minor axis $b$ , centre at the origin} whose two sides are parallel to two given straight lines, prove that the third side touches a fixed ellipse.(E$_2$)
Initially I assumed the fixed ellipse (E$_2$) to be standard one, the equation of sides given to me to be $$A(x-acos\theta)^2 +2H(x-acos\theta)(y-bsin\theta) +B(y-bsin\theta)^2=0$$, and the equation of the third side to be $$y=mx+c$$ Then I shifted the origin to a parametric coordinate of E$_1$ ,$(acos\theta,bsin\theta)$ and changed the equation of ellipse and the straight line accordingly. Now I homogenised the equation and compared the coefficients , but it gives a tedious equation of straight line. Is there an easier approach to this ? Or are there any guidelines to use the aforementioned approach. Thanks in advance.
All ellipses are affine images of the unit circle. Moreover, intersections, tangents and parallel lines are all affine properties, and the affine image of an ellipse is another ellipse, so w.l.o.g. we need only prove this for the unit circle.
If the given pair of lines are parallel, then the third side of the triangle collapses to a point and $E_2=E_1$, so we assume that the given lines aren’t parallel. Rotate the coordinate system so that the acute bisector of the lines is parallel to the $x$-axis and let $\theta$ be the acute angle between the lines. The two parallel lines through a point $A=(\cos t,\sin t)$ on the circle are $$x \sin{\frac\theta2} \pm y \cos{\frac\theta2} = \pm \sin\left(t-\frac\theta2\right).$$ The other two vertices of the triangle can be found by reflecting $A$ in the two perpendiculars through the origin of these two lines: $$B = (-\cos(t-\theta), \sin(t-\theta)) \\ C = (-\cos(t+\theta),\sin(t+\theta)).$$ The equation of the line through $B$ and $C$, after a bit of trigonometric manipulation, is $$2x\cos t\sin\theta - 2y\sin t\sin\theta + \sin{2\theta} = 0,$$ which can be further simplified to $$x \cos t - y \sin t + \cos\theta = 0$$ by dividing by $2\sin\theta$. These lines are at a constant distance of $\cos\theta$ from the origin, so are all tangent to the circle $x^2+y^2=\cos^2\theta$.