ઉપવલયના પ્રમાણિત સમીકરણ (y-અક્ષ પ્રત્યે) માં | vedclass

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Question

Let $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1 ; \mathrm{a}>\mathrm{b}$

$2 b=\frac{4}{\sqrt{3}} \Rightarrow

Vedclass · Accepted Answer

$\frac{1}{2} \sqrt{\frac{11}{3}}$

Vedclass · Answer

Let $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1 ; \mathrm{a}>\mathrm{b}$

$2 b=\frac{4}{\sqrt{3}} \Rightarrow b=\frac{2}{\sqrt{3}} \Rightarrow b^{2}=\frac{4}{3}$

tangent $\mathrm{y}=\frac{-\mathrm{x}}{6}+\frac{4}{3}$ compare with

$\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{a}^{2} \mathrm{m}^{2}+\mathrm{b}^{2}}$

$\Rightarrow \mathrm{m}=\frac{-1}{6} \Rightarrow \sqrt{\frac{\mathrm{a}^{2}}{36}+\frac{4}{3}}=\frac{4}{3} \Rightarrow \mathrm{a}=4$

$e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\frac{1}{2} \sqrt{\frac{11}{3}}$

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