Question

Find ecentricity of an ellipse whose latus rectum is 12.​

Answer 1

Answer:

12=2a(1-e^2)

6-a/a=-e^2

a-6/a=e^2

e=(a-6/a)^1/2

Answer 2

CORRECT QUESTION :

Find ecentricity of an ellipse whose latus rectum is 12 and the major axis is half of latus rectum

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Eccentricity=\iota}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{ \underline \bold{Given: }} \\ \tt{: \implies Latus \: rectum = 12} \\\\ \tt{:\implies Major\:axis=half\:of\:latus\:rectum} \\\\ \red{ \underline \bold{To \: Find: }} \\ \tt{: \implies Eccentricity =?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt{: \implies Latus \: rectum = \frac{2 {b}^{2} }{a} } \\ \\ \tt{: \implies 12= \frac{ {2b}^{2} }{a} } \\ \\ \tt{: \implies {b}^{2} = 6a} - - - - - (1)\\\\ \bold{As\:we\:know\:that}\\ \tt{:\implies major\:axis=6}\\\\ \tt{:\implies 2a=6}\\\\ \tt{:\implies a=3}$

$\bold{As \: we \: know \: that} \\ \tt{: \implies {b}^{2} = {a}^{2}(1 - {e}^{2} )} \\ \\ \text{Putting \: value \: of } {b}^{2} \\ \tt{: \implies 6a = {a}^{2} ( 1 - {e}^{2} ) }\\ \\ \tt{: \implies 6\times 3 = 9(1 - {e}^{2})} \\ \\ \tt{: \implies 1-{e}^{2} = 2} \\ \\ \tt{: \implies e = \sqrt{-1} } \\ \\ \green{\tt{: \implies e = \iota}}$