Question

For the ellipse x^2/5 + y^2/4 = 1
What will be the eccentricity ?

Plz help me.

Answer 1

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

$\green{\tt{\therefore{Eccentricity=\frac{1}{\sqrt{5}}}}}$

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

$\green{\underline \bold{Given :}} \\ \tt: \implies Eqn \: of \: ellipse \: \frac{ {x}^{2} }{5} + \frac{ {y}^{2} }{4} = 1 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Eccentricity = ?$

• According to given question :

$\tt: \implies \frac{ {x}^{2} }{5} + \frac{ {y}^{2} }{4} = 1 \\ \\ \tt: \implies \frac{ {x}^{2} }{ (\sqrt{5})^{2} } + \frac{ {y}^{2} }{ ({2})^{2} } = 1 \\ \\ \text{So, \: it \: is \: in \: the \: form \: of} \\ \tt: \implies \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1 \\ \\ \bold{Where : } \\ \tt \circ \: a^{2} = 5\\ \\ \tt \circ \: b^{2} = 4 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies {b}^{2} = {a}^{2} (1 - {e}^{2} ) \\ \\ \tt: \implies 4 = 5(1 - {e}^{2} ) \\ \\ \tt: \implies \frac{4}{5} = 1 - {e}^{2} \\ \\ \tt: \implies {e}^{2} =1 - \frac{4}{5} \\ \\ \tt: \implies {e}^{2} = \frac{1}{5} \\ \\ \tt: \implies e = \sqrt{ \frac{1}{5} } \\ \\ \green{\tt: \implies e = \frac{1}{ \sqrt{5} } }$

Answer 2

Given

An ellipse

$\frac{x^2}{5}$+$\frac{y^2}{4}$=1

To Find

Eccentricity

Solution

From given

$\frac{x^2}{5}$+$\frac{y^2}{4}$=1

1. a²=5
2. b²=4

$\boxed{\boxed{e^2=1-\frac{b^2}{a^2}}}$

Putting the values of 1 and 2 in E formula

$\boxed{\boxed{e^2=1-\frac{b^2}{a^2}}}$

Step by step calculation

=>$e^2=1-\frac{b^2}{a^2}$

=>$e^2=1-\frac{4}{5}$

=>$e^2=\frac{5-4}{5}$

=>$e^2=\frac{1}{5}$

=>e=$\sf{\dfrac{1}{\sqrt{5}}}$

Hence,the eccentricity is $\sf{\dfrac{1}{\sqrt{5}}}$