Question

Find the eccentricity of the locus of the incentre of triangle PS₂S₁, where e is the eccentricity of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and S₁, S₂ are its foci and P is any point on it.

• A. $\sqrt{\frac{2e}{1+e}}$
• B. $\sqrt{\frac{1-e}{1+e}}$
• C. $\sqrt{\frac{e}{1+e}}$
• D. $\sqrt{\frac{2e}{1-e}}$

Answer 1

Answer: (A)

$\sqrt{\frac{2e}{1+e}}$

Answer 2

The foci of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $S_1(ae, 0)$ and $S_2(-ae, 0)$. A point on the ellipse is $P(a \cos \theta, b \sin \theta)$.

The side lengths of triangle $PS_1S_2$ are $PS_1 = a - ae \cos \theta$, $PS_2 = a + ae \cos \theta$, and $S_1S_2 = 2ae$.

The incenter $(h, k)$ coordinates are derived using the incenter formula: $h = ae \cos \theta$ and $k = \frac{eb \sin \theta}{1+e}$.

Eliminating $\theta$ from $\cos \theta = h/(ae)$ and $\sin \theta = k(1+e)/(eb)$ using $\cos^2 \theta + \sin^2 \theta = 1$ gives the locus equation: $\frac{h^2}{(ae)^2} + \frac{k^2}{(eb/(1+e))^2} = 1$.

This is an ellipse with semi-major axis $A = ae$ and semi-minor axis $B = \frac{eb}{1+e}$. Its eccentricity $e'$ is calculated as $e'^2 = 1 - \frac{B^2}{A^2} = 1 - \frac{(eb/(1+e))^2}{(ae)^2} = 1 - \frac{b^2}{a^2(1+e)^2}$. Substituting $b^2 = a^2(1-e^2)$ and simplifying yields $e'^2 = 1 - \frac{a^2(1-e^2)}{a^2(1+e)^2} = \frac{(1+e)^2 - (1-e^2)}{(1+e)^2} = \frac{1+2e+e^2-1+e^2}{(1+e)^2} = \frac{2e+2e^2}{(1+e)^2} = \frac{2e(1+e)}{(1+e)^2} = \frac{2e}{1+e}$. Thus, the eccentricity of the locus of the incenter is $e' = \sqrt{\frac{2e}{1+e}}$.