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Sagot :
The equation of the ellipse having foci (±2,0) and vertices (±5,0) is [tex]x^{2} /4+y^{2} /21=1[/tex].
Given the foci of ellipse is (±2,0) and vertices (±5,0).
We are required to find the equation of ellipse having foci (±2,0) and vertices (±5,0).
Equation is the relationship between two or more variables that are expressed in equal to form. Equation of two variables looks like ax+by=c.
Equation of ellipse is as under:
[tex]x^{2} /a^{2} +y^{2} /b^{2} =1[/tex],where a is the semi major axis.
We have been given that a=5, c=2.
We know that [tex]c^{2} =a^{2} -b^{2}[/tex]
We have to find the value of b.
b=[tex]\sqrt{a^{2} -c^{2} }[/tex]
=[tex]\sqrt{5^{2} -2^{2} }[/tex]
=[tex]\sqrt{25-4}[/tex]
=[tex]\sqrt{21}[/tex]
Using the values of a and b in the equation [tex]x^{2} /a^{2} +y^{2} /b^{2} =1[/tex]
[tex]x^{2} /(2)^{2} +y^{2}/\sqrt{21} ^{2} =1[/tex]
[tex]x^{2} /4+y^{2} /21=1[/tex]
Hence the equation of the ellipse having foci (±2,0) and vertices (±5,0) is [tex]x^{2} /4+y^{2} /21=1[/tex].
Learn more about ellipse at https://brainly.com/question/16904744
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