Question

The distance between two foci of eclipse 5x²+4y² = 20 is:​

Answer 1

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dividing both sides by 20 ;

$\rightarrow \: { \tt{\frac{5x {}^{2} }{20} + \frac{4y {}^{2} }{20} = \frac{20}{20} }} \\ \\ \rightarrow \:{ \tt{ \frac{x {}^{2} }{4} + \frac{y {}^{2} }{5} = 1}} \\ \\ { \tt{a = 4 \: and \: b = 5}}$

hence, by comparison distance between two foci is 5+5 = 10 units.

Answer 2

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dividing both sides by 20 ;

$\begin{lgathered}\rightarrow \: { \tt{\frac{5x {}^{2} }{20} + \frac{4y {}^{2} }{20} = \frac{20}{20} }} \\ \\ \rightarrow \:{ \tt{ \frac{x {}^{2} }{4} + \frac{y {}^{2} }{5} = 1}} \\ \\ { \tt{a = 4 \: and \: b = 5}}\end{lgathered}$

→205x2+204y2=2020→4x2+5y2=1a

=4andb=5

hence, by comparison distance between two foci is 5+5 = 10 units