Question

Let these base $AB$ of a triangle $ABC$ be fixed and the vertex $C$ lie on a fixed circle of radius $r$. Lines through $A$ and $B$ are drawn to intersect $CB$ and $CA$, respectively, at $E$ and $F$ such that $CE:EB=1:1$ and $CF:FA=1:2$. If the point of intersection $P$ of these lines lies on the median through $AB$ for all positions of $AB$ then the focus of $P$ is

• A. a circle of radius $\frac{r}{2}$
• B. a circle of radius $2r$
• C. a parabola of latus rectum $4r$
• D. a rectangular hyperbola

Answer 1

The correct option is A a circle of radius $\frac{r}{2}$

Let A and B be $(-a,0)$ and $(a,0)$. Also let P be $(h,k)$.
Then by geometry, we know $\frac{CP}{PO}=\frac{CF}{FA}+\frac{CE}{EB}$
$\therefore \frac{CP}{PO}=1$
If $C(\alpha ,\beta )$ lies on $x^2+y^2+2gx+2fy+c=0$, then $\alpha =2h$ and $\beta =2k$
$\Rightarrow 4(h^2+k^2+gh+fk)+c=0$
$\therefore$ Locus of P(h, k) is $x^2+y^2+gx+fy+\frac{c}{4}=0$ which is a circle of radius $=\sqrt{{\left(\frac{g}{2}\right)}^2+{\left(\frac{f}{2}\right)}^2-\frac{c}{4}}$
$=\frac{1}{2}\sqrt{g^2+f^2-c}$
$=\frac{r}{2}$