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.
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.