Question

Let $P$ be a point on the ellipse $x^2 + 2y^2 = 2$ with foci $S$ & $S'$. The locus of incentre of $\triangle PSS'$ is a conic $C$.

Let $\beta$ equal to length of the latus rectum of the conic $C$. $[100\beta]$ is equal to, where $[x]$ is greatest integer less than or equal to $x$.

Answer 1

34

Answer 2

Solution:

1. The given ellipse is $x^2 + 2y^2 = 2 \quad\Leftrightarrow\quad \frac{x^2}{2} + \frac{y^2}{1} = 1,$ with foci $S(1,0)$ and $S'(-1,0)$.

2. For any point $P=(x,y)$ on the ellipse, the incenter $I=(h,k)$ of $\triangle PSS'$ is found (via a barycentric/incenter formula) to be related to $P$ by:

   $$x = h\sqrt{2},\quad y = k(1+\sqrt{2}).$$

3. Substituting into the ellipse equation:

   $$(h\sqrt{2})^2 + 2\big[k(1+\sqrt{2})\big]^2 = 2 \quad\Longrightarrow\quad 2h^2 + 2(1+\sqrt{2})^2 k^2 = 2.$$

   Dividing by 2 gives the locus of $I$:

   $$h^2 + (1+\sqrt{2})^2 k^2 = 1.$$

   This represents an ellipse in $h,k$ where if written in standard form

   $$\frac{h^2}{1^2}+\frac{k^2}{\left(\frac{1}{1+\sqrt{2}}\right)^2}=1,$$

   we identify $a=1$ and $b=\frac{1}{1+\sqrt{2}}$.

4. The latus rectum $L$ of an ellipse with parameters $a, b$ is given by

   $$L = \frac{2b^2}{a}.$$

   Here,

   $$\beta = L = \frac{2}{(1+\sqrt{2})^2}.$$

5. Notice that

   $$(1+\sqrt{2})^2 = 3+2\sqrt{2}.$$

   Thus,

   $$\beta = \frac{2}{3+2\sqrt{2}} = 2(3-2\sqrt{2}) \quad \text{(after rationalization)}.$$

   A simpler calculation is to compute numerically:

   $$1+\sqrt{2} \approx 2.414,\quad (2.414)^2 \approx 5.828,$$

   so

   $$\beta \approx \frac{2}{5.828} \approx 0.343.$$

   Then,

   $$100\beta \approx 34.3.$$

   Taking the floor function, we have $[100\beta] = 34$.