Question

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

• A. $(32/9,22/9)$
• B. $(23/9,2/9)$
• C. $(34/9,11/9)$
• D. none of these

Answer 1

Answer: (C)

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 $S^{\prime \prime }\equiv (6,3)$
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 $PS+P{S}^{\prime }=2a$
Since $P{S}^{\prime }=P{S}^{\prime \prime }$, there exists one and only one point $P$ on
$L=0$ such that PS $+P{S}^{\prime \prime }=2a$.
Hence, P should be the collinear with SS".
Hence, $P$ is a point of intersection of $S{S}^{\prime \prime }(4x-5y=9)$
and $x+y-5=0$, and $P\equiv (34/9,11/9)$.