Question

Show that the locus of the incentre of the variable triangle $PF_2F_1$ is an

ellipse whose eccentricity is $\sqrt{\frac{2e}{1+e}}$ where $e$ is the eccentricity of standard

ellipse when $P$ moves on a standard ellipse.

Answer 1

ellipse whose eccentricity is $\sqrt{\frac{2e}{1+e}}$

Answer 2

Let the standard ellipse be given by the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ is the semi-major axis and $b$ is the semi-minor axis. The eccentricity of this ellipse is $e$, related by $b^2 = a^2(1-e^2)$.

The foci of the ellipse are $F_1 = (ae, 0)$ and $F_2 = (-ae, 0)$. Let $P = (a \cos \theta, b \sin \theta)$ be a point on the ellipse.

We are considering the triangle $PF_2F_1$. The side lengths of this triangle are:

1. $PF_1 = r_1 = a - ae \cos \theta$ (distance from a point on the ellipse to the focus $F_1$)
2. $PF_2 = r_2 = a + ae \cos \theta$ (distance from a point on the ellipse to the focus $F_2$)
3. $F_1F_2 = 2ae$ (distance between the foci)

The sum of the distances from $P$ to the foci is $r_1 + r_2 = (a - ae \cos \theta) + (a + ae \cos \theta) = 2a$. The perimeter of the triangle $PF_2F_1$ is $S = r_1 + r_2 + F_1F_2 = 2a + 2ae = 2a(1+e)$.

Let $I = (h, k)$ be the incenter of the triangle $PF_2F_1$. The coordinates of the vertices are $P(a \cos \theta, b \sin \theta)$, $F_1(ae, 0)$, and $F_2(-ae, 0)$. Using the incenter formula $(h, k) = \left(\frac{r_2 x_{F_1} + r_1 x_{F_2} + F_1F_2 x_P}{S}, \frac{r_2 y_{F_1} + r_1 y_{F_2} + F_1F_2 y_P}{S}\right)$:

For the x-coordinate ($h$): $h = \frac{(a + ae \cos \theta)(ae) + (a - ae \cos \theta)(-ae) + (2ae)(a \cos \theta)}{2a(1+e)}$ $h = \frac{a^2e + a^2e^2 \cos \theta - a^2e + a^2e^2 \cos \theta + 2a^2e \cos \theta}{2a(1+e)}$ $h = \frac{2a^2e^2 \cos \theta + 2a^2e \cos \theta}{2a(1+e)}$ $h = \frac{2a^2e \cos \theta (e+1)}{2a(1+e)}$ $h = ae \cos \theta$

For the y-coordinate ($k$): $k = \frac{(a + ae \cos \theta)(0) + (a - ae \cos \theta)(0) + (2ae)(b \sin \theta)}{2a(1+e)}$ $k = \frac{2aeb \sin \theta}{2a(1+e)}$ $k = \frac{eb \sin \theta}{1+e}$

Now we have parametric equations for the locus of the incenter $(h, k)$:

1. $\cos \theta = \frac{h}{ae}$
2. $\sin \theta = \frac{k(1+e)}{eb}$

Using the identity $\cos^2 \theta + \sin^2 \theta = 1$: $\left(\frac{h}{ae}\right)^2 + \left(\frac{k(1+e)}{eb}\right)^2 = 1$ $\frac{h^2}{(ae)^2} + \frac{k^2}{\left(\frac{eb}{1+e}\right)^2} = 1$

This is the equation of an ellipse of the form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$, where the semi-axes are $A = ae$ and $B = \frac{eb}{1+e}$.

To find the eccentricity of this new ellipse, we need to determine which axis is the major axis. We compare $A^2$ and $B^2$: $A^2 = (ae)^2 = a^2e^2$ $B^2 = \left(\frac{eb}{1+e}\right)^2 = \frac{e^2b^2}{(1+e)^2}$ Substitute $b^2 = a^2(1-e^2)$: $B^2 = \frac{e^2a^2(1-e^2)}{(1+e)^2}$

Now, compare $A^2$ and $B^2$: $A^2 - B^2 = a^2e^2 - \frac{e^2a^2(1-e^2)}{(1+e)^2}$ $A^2 - B^2 = a^2e^2 \left(1 - \frac{1-e^2}{(1+e)^2}\right)$ $A^2 - B^2 = a^2e^2 \left(\frac{(1+e)^2 - (1-e^2)}{(1+e)^2}\right)$ $A^2 - B^2 = a^2e^2 \left(\frac{1+2e+e^2 - 1+e^2}{(1+e)^2}\right)$ $A^2 - B^2 = a^2e^2 \left(\frac{2e+2e^2}{(1+e)^2}\right)$ $A^2 - B^2 = a^2e^2 \left(\frac{2e(1+e)}{(1+e)^2}\right)$ $A^2 - B^2 = a^2e^2 \left(\frac{2e}{1+e}\right)$

Since $0 < e < 1$, $a^2e^2 > 0$ and $\frac{2e}{1+e} > 0$. Therefore, $A^2 - B^2 > 0$, which means $A > B$. So, the major axis of the new ellipse is along the x-axis, and its semi-major axis is $A = ae$, and its semi-minor axis is $B = \frac{eb}{1+e}$.

The eccentricity of the new ellipse, let's call it $e'$, is given by $e'^2 = 1 - \frac{B^2}{A^2}$. $e'^2 = 1 - \frac{\frac{e^2b^2}{(1+e)^2}}{a^2e^2}$ $e'^2 = 1 - \frac{b^2}{a^2(1+e)^2}$ Substitute $b^2 = a^2(1-e^2)$: $e'^2 = 1 - \frac{a^2(1-e^2)}{a^2(1+e)^2}$ $e'^2 = 1 - \frac{1-e^2}{(1+e)^2}$ $e'^2 = \frac{(1+e)^2 - (1-e^2)}{(1+e)^2}$ $e'^2 = \frac{1+2e+e^2 - 1+e^2}{(1+e)^2}$ $e'^2 = \frac{2e+2e^2}{(1+e)^2}$ $e'^2 = \frac{2e(1+e)}{(1+e)^2}$ $e'^2 = \frac{2e}{1+e}$

Thus, the eccentricity of the locus of the incenter is $e' = \sqrt{\frac{2e}{1+e}}$. The locus is an ellipse.