analytic geometryconic sections
Find the equation of the focal chord of the
ellipse $3x^2 + 4y^2 = 48$ , whose length is 7.
I found that one of the foci of the ellipse is (2; 0). If I express the equation of the line L that is requested as L: y = mx + b, and replace the coordinates of the point (2; 0), I obtain b = -2m. With this we have L: y = m (x-2). I only need the slope m to solve the problem, but I don't know how to get it. Tried plugging into ellipse equation but got nowhere.
Part I:
After parametrization $P(t)\equiv(ae+r\cos t,r\sin t)$: $$\frac{(ae+r\cos t)^2}{a^2}+\frac{r^2\sin^2t}{b^2}=1$$ $$b^2a^2e^2+b^2r^2\cos^2t+2ab^2er\cos t+a^2r^2\sin^2t-a^2b^2=0\\ (a^2\sin^2t+b^2\cos^2t)r^2+(2ab^2e\cos t)r+(b^2a^2e^2-a^2b^2)=0\\ (a^2\sin^2t+b^2\cos^2t)r^2+(2ab^2e\cos t)r-b^4=0\quad(\color{red}{b^2=a^2(1-e^2)})$$ Now: $$r_1=l_1,r_2=-l_2\implies |l_1+l_2|=|r_1-r_2|=\left|\sqrt{(r_1+r_2)^2-4r_1r_2}\right|$$ $$|l_1+l_2| =\sqrt{\left(\frac{2abe\cos t}{(a^2\sin^2t+b^2\cos^2t)}\right)^2+4\frac{(b^4)}{(a^2\sin^2t+b^2\cos^2t)}}\\=\sqrt{\frac{4a^2b^4e^2\cos^2t}{(a^2\sin^2t+b^2\cos^2t)^2}+4\frac{(b^4)}{(a^2\sin^2t+b^2\cos^2t)}}\\ =\frac{2b^2}{(a^2\sin^2t+b^2\cos^2t)}\sqrt{a^2e^2\cos^2t+a^2\sin^2t+\underbrace{b^2}_{b^2=a^2(1-e^2)}\cos^2t}\\ =\frac{2b^2}{(a^2\sin^2t+b^2\cos^2t)}\sqrt{a^2e^2\cos^2t+a^2\sin^2t+a^2\cos^2t-a^2e^2cos^2t}\\ \fbox{$\huge PQ=\frac{2ab^2}{(a^2\sin^2t+b^2\cos^2t)}$} $$
Part II
$$(a^2\sin^2t+b^2\cos^2t)r^2+(2ab^2e\cos t)r-b^4=0$$
Now: $$\frac{PS}{SQ}=\frac{l_1}{l_2}=\left|\frac{r_1}{r_2}\right|=\left|\frac{(r_1+r_2)+(r_1-r_2)}{(r_1+r_2)-(r_1-r_2)}\right|=\left|\frac{-\frac{(2ab^2e\cos t)}{(a^2\sin^2t+b^2\cos^2t)}+\frac{2ab^2}{(a^2\sin^2t+b^2\cos^2t)}}{-\frac{(2ab^2e\cos t)}{(a^2\sin^2t+b^2\cos^2t)}-\frac{2ab^2}{(a^2\sin^2t+b^2\cos^2t)}}\right|=\frac{1-e\cos t}{1+e\cos t}$$ Similiarly: $$\frac{PS'}{S'R}=\frac{1+e\cos u}{1-e\cos u}$$ Now take a triangle with vertices $P,S$ and the foot of perpendicular from P to Major Axis, then: $$\cos t=\frac{a\cos\theta-ae}{\sqrt{(a\cos\theta-ae)^2+b^2\sin^2\theta}}=\frac{\cos\theta-e}{1-e\cos\theta}$$ Similiarly: $$\cos u=\frac{\cos\theta+e}{1+e\cos\theta}$$ Now: $$\frac{PS}{SQ}=\frac{1-e\cos t}{1+e\cos t}=\frac{1-e\cos\theta-e(\cos\theta-e)}{1-e\cos\theta+e(\cos\theta-e)}=\frac{1-2e\cos\theta+e^2}{1-e^2}$$ Similiarly: $$\frac{PS'}{S'R}=\frac{1+2e\cos\theta+e^2}{1-e^2}$$ So: $$\frac{PS}{SQ}+\frac{PS'}{S'R}=\frac{1-2e\cos\theta+e^2}{1-e^2}+\frac{1+2e\cos\theta+e^2}{1-e^2}\\\fbox{$\huge\frac{PS}{SQ}+\frac{PS'}{S'R}=2\left(\frac{1+e^2}{1-e^2}\right)$}$$
This exercise can be understood as an application of a general result about perimeter bisectors of triangles.
Proposition. Given $\triangle ABC$ with incircle $\bigcirc I$ meeting the edges at $D$, $E$, $F$ as shown. If $F^\prime$ is the point opposite $F$ in $\bigcirc I$, and if $F^{\prime\prime}$ is the point where $\overleftrightarrow{CF^\prime}$ meets $\overline{AB}$, then $$|\overline{CA}|+|\overline{AF^{\prime\prime}}| = |\overline{CB}|+|\overline{BF^{\prime\prime}}| \tag{$\star$}$$ so that $\overline{CF^{\prime\prime}}$ is a perimeter bisector of $\triangle ABC$.
Proof of Proposition. Let the perpendicular to $\overline{FF^\prime}$ at $F^\prime$ meet the edges of the triangle at $A^\prime$ and $B^\prime$. By tangent properties of circles, we have $$\overline{CE}\cong\overline{CD} \qquad \overline{A^\prime E}\cong\overline{A^\prime F^\prime} \qquad \overline{B^\prime D}\cong\overline{B^\prime F^\prime}$$ Consequently, $|\overline{CA^\prime}| + |\overline{A^\prime F^\prime}| = |\overline{CB^\prime}| + |\overline{B^\prime F^\prime}|$, so that $\overline{CF}$ is a perimeter bisector of $\triangle A^\prime B^\prime C$. The Proposition holds by the similarity of $\triangle ABC$ and $\triangle A^\prime B^\prime C$. $\square$
The Proposition has a helpful corollary.
Corollary. Given $\triangle ABC$ with incenter $I$ and perimeter bisector $\overline{CF^{\prime\prime}}$, if $M$ is on $\overline{AB}$ such that $\overline{IM} \parallel \overline{CF^{\prime\prime}}$, then $M$ is the midpoint of $\overline{AB}$.
Proof of Corollary. The points of tangency of the triangle with its incircle separate the perimeter into three pairs of congruent segments, marked $a$, $b$, $c$. Thus, the semi-perimeter of $\triangle ABC$ is $a+b+c$, and since $|\overline{BC}| = b+c$, it follows that $|\overline{BF^{\prime\prime}}| = a = |\overline{AF}|$. Thus, $\overline{FF^{\prime\prime}}$ lies between congruent segments. In $\triangle FF^\prime F^{\prime\prime}$, segment $\overline{IM}$ passes through the midpoint of one side ($\overline{FF^\prime}$) and is parallel to another ($\overline{F^\prime F^{\prime\prime}}$); it necessarily meets the third side ($\overline{FF^{\prime\prime}}$) at its midpoint, which must also be the midpoint of $\overline{AB}$. $\square$
To solve the original problem, it basically suffices to embed the above triangle into an ellipse:
In the above, the ellipse's foci are $C$ and $F^{\prime\prime}$, and $\overline{AB}$ is a chord through the latter. The fundamental nature of ellipses implies that $(\star)$ holds; therefore, $\overline{CF^{\prime\prime}}$ is a perimeter bisector of $\triangle ABC$. Moreover, the reflection property of ellipses implies that normals at $A$ and $B$ bisect angles $\angle CAF^{\prime\prime}$ and $\angle CBF^{\prime\prime}$; therefore, the intersection of these normals is the incenter of $\triangle ABC$. The result follows by the Corollary. $\square$
Best Answer
The equation of the ellipse in standard form is
$\dfrac{x^2}{16} + \dfrac{y^2}{12} = 1 $
i.e. $a^2 = 16 , b^2 = 12 $
So the eccentricity is $e = \sqrt{1 - \dfrac{12}{16}} = \dfrac{1}{2}$.
Hence the foci are at $(- a e, 0 )$ and $(a e, 0)$, i.e. at $(-2, 0)$ and $(2, 0)$.
Now pass a line making an angle $\theta $ with the positive $x$ axis, through the right focus, then its parameteric equation is
$ p(t) = (2, 0) + t (\cos \theta , \sin \theta)$
Place this point on the ellipse, then
$(2 + t \cos \theta)^2 / 16 + (t \sin \theta)^2/12 = 1$
Expanding we get a quadratic equation in $t$ as follows
$ a_2 t^2 + a_1 t + a_0 = 0 $
Where $a_2 = \dfrac{\cos^2 \theta}{16} + \dfrac{\sin^2 \theta } { 12 }, a_1 = \dfrac{1}{4} \cos \theta , a_0 = -\dfrac{3}{4} $
The discriminant is
$D= a_1^2 - 4 a_2 a_0 = \dfrac{1}{4}$
Thus the difference in the two solutions is
$\Delta t = \dfrac{\sqrt{\dfrac{1}{4}}}{ a_2 } = 7 $
Hence,
$\dfrac{1}{2} = \dfrac{7}{16} \cos^2 \theta +\dfrac{7}{12} \sin^2 \theta $
Using $\cos^2 \theta = \dfrac{1}{2}(1 + \cos 2 \theta) $, and $ \sin^2 \theta = \dfrac{1}{2} (1 - \cos 2 \theta )$, the above equation becomes,
$\dfrac{1}{2} = \dfrac{49}{96} - \dfrac{7}{96} \cos 2 \theta $
which has the solutions
$ \theta = \pm \dfrac{1}{2}\cos^{-1} \dfrac{1}{7} $
Now the equation of the focal chord is fully known.