Find-the-ecentricity-If-1-lactus-rectum-is-half-major-axis-2-lactus-rectum-is-half-minor-axis-

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

F i n d t h e e c e n t r i c i t y I f

(1) l a c t u s r e c t u m i s h a l f

m a j o r a x i s

(2) l a c t u s r e c t u m i s h a l f

m i n o r a x i s

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

1) e = \frac{c}{a}

L T \frac{2 b^{2}}{a} = a 2 b^{2} = a^{2}

c^{2} = a^{2} - b^{2} \to f o r e l i p s e

= a^{2} - \frac{a^{2}}{2}

= \frac{a^{2}}{2} c = \pm \frac{a}{\sqrt{2}}

s o e = \frac{c}{a} = \pm \frac{a}{\sqrt{2} \times a} = \pm \frac{1}{\sqrt{2}}

f o r h y p e r b o l a

c^{2} = a^{2} + b^{2}

c^{2} = a^{2} + \frac{a^{2}}{2} = \frac{3 a^{2}}{2}

c = \pm \sqrt{\frac{3}{2}} a

e = \frac{c}{a} = \pm \sqrt{\frac{3}{2}}

2) \frac{2 b^{2}}{a} = b

a = 2 b

c^{2} = a^{2} - b^{2} f o r e l l i p s e

c^{2} = a^{2} - \frac{a^{2}}{4} = \frac{3 a^{2}}{4}

e = \frac{c}{a} = \pm \frac{\sqrt{3}}{2}

f o r h y p e r b o l a

c^{2} = a^{2} + b^{2}

c^{2} = a^{2} + \frac{a^{2}}{4}

c^{2} = \frac{5 a^{2}}{4}

e = \frac{c}{a} = \pm \frac{\sqrt{5}}{2}

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

s i r t h e a n s g i v e n a r e

(1) e = \pm \frac{\sqrt{2}}{2}

(2) e = \pm \frac{\sqrt{3}}{2}

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

o k i h a v e i g o n r e d \pm s i g n \dots l e t m e a d d \pm s i g n