Thank you for reporting, we will resolve it shortly
Q.
An ellipse has points $(1,-1)$ and $(2,-1)$ as its foci and $x+y-5=0$ as one of its tangents. Then the point where this line touches the ellipse is
Conic Sections
Solution:
The equation has three roots. Hence, there normals can be drawn.
Let the image of S' be with respect to $x+y-5=0$. Then,
$\frac{ h -2}{1}=\frac{ k +1}{1}=\frac{-2(-4)}{2} $
or $S'' \equiv(6,3)$
Let $P$ be the point of contact.
Because the line $L =0$ is tangent to the ellipse, there exists a point $P$ uniquely on the line such that $PS + PS '=2 a$
Since $PS'= PS''$, there exists one and only one point $P$ on
$L=0$ such that PS $+P S''=2 a$.
Hence, P should be the collinear with SS".
Hence, $P$ is a point of intersection of $S S''(4 x-5 y=9)$
and $x+y-5=0$, and $P \equiv(34 / 9,11 / 9)$.
