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  • If radii of director circles of <img alt="\mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Cfrac%7Bx%5E2%7D%7Ba%5E2%7D&plus;%5Cfrac

If radii of director circles of \mathrm{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1} and \mathrm{\frac{x^2}{a^2}-\frac{y^2}{b_1^2}=1} are 2r and r respectively and \mathrm{e_e} and \mathrm{e_h} be the eccentricities of the ellipse and hyperbola respectively, then

Option: 1

\mathrm{2 e_h^2-e_c^2=6}


Option: 2

\mathrm{e_e^2-4 e_h^2=6}


Option: 3

\mathrm{4 e_h^2-e_e^2=6}


Option: 4

\mathrm{\text { None of these }}


Answers (1)

best_answer

The equations of the director circles of ellipse and hyperbola are x^2+y^2=a^2+b^2\, \, and \, \, x^2+y^2=a^2-b^2{ }_1 respectively.

\mathrm{ \begin{aligned} & \therefore 2 r=\sqrt{a^2+b^2} \text { and } r=\sqrt{a^2-b_1^2} \\\\ & \Rightarrow 2 \sqrt{a^2-b_1^2}=\sqrt{a^2+b^2} \Rightarrow 4\left(a^2-b_1^2\right)=a^2+b^2 \\\\ & \Rightarrow 4\left(1-\frac{b_1^2}{a^2}\right)=\left(1+\frac{b^2}{a^2}\right) \\\\ & \Rightarrow 4\left\{1-\left(e_h^2-1\right)\right\}=1+\left(1-e_e^2\right) \\\\ & \Rightarrow 8-4 e_h^2=2-e_e^2 \Rightarrow 4 e_h^2-e_e^2=6 \end{aligned} }

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HARSH KANKARIA

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