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  • A variable point P on an ellipse of eccentricity is joined to its foci <img alt="\mathrm{ S, S^{\prime}}" sr

A variable point P on an ellipse of eccentricity 3 / 5 is joined to its foci \mathrm{ S, S^{\prime}}, then the locus of the incentre of the \mathrm{\triangle P S S^{\prime}} is an ellipse. If \mathrm{\lambda} be the eccentricity of the ellipse then the value of \mathrm{128 \lambda^2} must be

Option: 1

98


Option: 2

96


Option: 3

92


Option: 4

91


Answers (1)

best_answer

Let the given ellipse be

\mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 }

Let the coordinates of P are \mathrm{(a \cos \phi, b \sin \phi)}
By hypothesis,

\mathrm{ \begin{aligned} & b^2=a^2\left(1-e^2\right) \text { and } S(a e, 0), S^{\prime}(-a e, 0) \\\\ & \therefore \quad S P=\text { focal distance of the point } P \end{aligned} }

\mathrm{ =a-a e \cos \phi }
and 

\mathrm{S^{\prime} P=a+a e \cos \phi}
Also, \mathrm{S^{\prime}=2 a e}
If (x, y) be the incentre of the \mathrm{\triangle P S S^{\prime}}, then

\mathrm{ \therefore x=\frac{(2 a e) a \cos \phi+a(1-e \cos \phi)(-a e)+a(1+e \cos \phi) a e}{2 a e+a(1-e \cos \phi)+a(1+e \cos \phi)} }
\mathrm{ x=a e \cos \phi }                                                                                                         ...(i)

and \mathrm{y=\frac{2 a e(b \sin \phi)+a(1+e \cos \phi) \cdot 0+a(1-e \cos \phi) \cdot 0}{2 a e+a(1-e \cos \phi)+a(1+e \cos \phi)}}

\mathrm{\Rightarrow y=\frac{e b \sin \phi}{(e+1)}}                                                           ...(ii)

Eliminating \phi from Eqs. (i) and (ii), we get

\mathrm{ \frac{x^2}{a^2 e^2}+\frac{y^2}{\left[\frac{b e}{e+1}\right]^2}=1 }
which represent an ellipse.
\mathrm{\because \quad \lambda} be its eccentricity.

\mathrm{ \begin{aligned} & \therefore \frac{b^2 e^2}{(e+1)^2}=a^2 e^2\left(1-\lambda^2\right) \\\\ & \Rightarrow \lambda^2=1-\frac{b^2}{a^2(e+1)^2} \\\\ & \Rightarrow \lambda^2=1-\frac{1-e^2}{(e+1)^2} \\\\ & \Rightarrow \lambda^2=1-\frac{1-e}{1+e}=\frac{2 e}{1+e} \\\\ & \Rightarrow \lambda=\sqrt{\left(\frac{2 e}{1+e}\right)}=\sqrt{\left(\frac{2 \times 3 / 5}{1+3 / 5}\right)}=\sqrt{\frac{3}{4}} \\\\ & \therefore 128 \lambda^2=128 \times \frac{3}{4} \end{aligned} }

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Rishi

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