Question
If a point $\mathrm{P}$ on the ellipse $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ is joined to $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ (foci of the ellipse) then,(i) show that the incentre of the triangle $\mathrm{PS}_{1} \mathrm{~S}_{2}$ lies on another ellipse.(ii) find the eccentricity of this ellipse in terms of the eccentricity of the given ellipse.
Step 1
The coordinates of $S_{1}$ and $S_{2}$ are $(0,0)$ and $(-ae,0)$ respectively. Let $P$ be a point on the ellipse with coordinates $(a\cos\theta, b\sin\theta)$. Show more…
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Key Concepts
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Show that the ellipse $ x^2/a^2 + y^2/b^2 = 1 $ and the hyperbola $ x^2/A^2 - y^2/B^2 = 1 $ are orthogonal trajectories if $ A^2 < a^2 $ and $ a^2 - b^2 = A^2 + B^2 $ (so the ellipse and hyperbola have the same foci)
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If $a>b>0,$ then the eccentricity of the ellipse $$\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1 \quad \text { or } \quad \frac{(x-h)^{2}}{b^{2}}+\frac{(y-k)^{2}}{a^{2}}=1$$ is the number $\frac{\sqrt{a^{2}-b^{2}}}{a} .$ Find the eccentricity of the ellipse whose equation is given. $$\frac{x^{2}}{18}+\frac{y^{2}}{25}=1$$
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Circles and Ellipses
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