Question

The eccentricity of the ellipse 9x2 + 25y2 = 225
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Answer 1

Step-by-step explanation:

9×2 +25y^2 =225

18 + 25y^2 =225

25y^2 =225 -18

25y^2 = 207

y^2 =207/25

y= (207/25)^1/2

y = 3√23/5

Answer 2

Given :

• Equation of ellipse : 9x² + 25y² = 225

To find :

• Eccentricity of the ellipse

Formula used :

$\sf equation \: of \: ellipse : \dfrac{ {x}^{2} }{ {a}^{2} } + \dfrac{ {y}^{2} }{ {b}^{2} } = 1$

Now eccentricity :-

$\sf \dfrac{ {b}^{2} }{{a}^{2}} = 1 - {e}^{2}$

Where e = eccentricity

Solution :

$\implies \sf 9 {x}^{2} + 25 {y}^{2} = 225$

$\implies \sf \dfrac{9}{225} {x}^{2} + \dfrac{25}{225} {y}^{2} = 1$

$\implies \sf \dfrac{{x}^{2}}{ (\frac{225}{9} )} + \dfrac{{y}^{2}}{( \frac{225}{25}) } = 1$

$\implies \sf \dfrac{{x}^{2}}{ {(\frac{15}{3} )}^{2} } + \dfrac{{y}^{2}}{( {\frac{15}{5})}^{2} } = 1$

$\implies \sf \dfrac{{x}^{2}}{ {(5)}^{2} } + \dfrac{{y}^{2}}{( {3)}^{2} } = 1$

$\implies \sf \dfrac{{x}^{2}}{ 25} + \dfrac{{y}^{2}}{ 9 } = 1$

Now, a = 5 and a²= 25

b = 3 and b² = 9

Now to find out eccentricity put the value of a² and b² :-

$\implies \sf \dfrac{ {b}^{2} }{{a}^{2}} = 1 - {e}^{2}$

$\implies \sf \dfrac{ 9 }{25} = 1 - {e}^{2}$

$\implies \sf 1 - {e}^{2} = \dfrac{ 9 }{25}$

$\implies \sf - {e}^{2} =\dfrac{ 9 }{25} - 1$

$\implies \sf - {e}^{2} =\dfrac{ 9 - 25}{25}$

$\implies \sf - {e}^{2} =\dfrac{ - 16}{25}$

$\implies \sf {e} = \sqrt{\dfrac{ 16}{25} }$

$\implies \sf e = \pm \dfrac{ 4}{5}$

But we have to neglect -4/5 because eccentricity of ellipse is always between zero and one.

$\implies \sf {e} = \dfrac{4}{5}$

Answer :

• The eccentricity of the ellipse 9x2 + 25y2 = 225= 4/5