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Question Number 51269 by peter frank last updated on 25/Dec/18

$$FindtheecentricityIf$$$$\left(1\right)lactusrectumishalf$$$$majoraxis$$$$\left(2\right)lactusrectumishalf$$$$minoraxis$$

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18

$$1)e=\frac{c}{a}$$$$LT\frac{2b^2}{a}=a2b^2=a^2$$$$$$$$c^2=a^2-b^2\to forelipse$$$$=a^2-\frac{a^2}{2}$$$$=\frac{a^2}{2}c=\pm \frac{a}{\sqrt{2}}$$$$soe=\frac{c}{a}=\pm \frac{a}{\sqrt{2}\times a}=\pm \frac{1}{\sqrt{2}}$$$$forhyperbola$$$$c^2=a^2+b^2$$$$c^2=a^2+\frac{a^2}{2}=\frac{3a^2}{2}$$$$c=\pm \sqrt{\frac{3}{2}}a$$$$e=\frac{c}{a}=\pm \sqrt{\frac{3}{2}}$$$$2)\frac{2b^2}{a}=b$$$$a=2b$$$$c^2=a^2-b^{2}forellipse$$$$c^2=a^2-\frac{a^2}{4}=\frac{3a^2}{4}$$$$e=\frac{c}{a}=\pm \frac{\sqrt{3}}{2}$$$$forhyperbola$$$$c^2=a^2+b^2$$$$c^2=a^2+\frac{a^2}{4}$$$$c^2=\frac{5a^2}{4}$$$$e=\frac{c}{a}=\pm \frac{\sqrt{5}}{2}$$$$$$

Commented by peter frank last updated on 25/Dec/18

$$sirtheansgivenare$$$$\left(1\right)e=\pm \frac{\sqrt{2}}{2}$$$$\left(2\right)e=\pm \frac{\sqrt{3}}{2}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 25/Dec/18

$$okihaveigonred\pm sign...letmeadd\pm sign$$

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