Question

Find an equation for the conic that satisfies the given conditions. ellipse, foci (±2, 0), vertices (±5, 0)

Answer 1

The equation of the ellipse having foci  (±2,0) and vertices (±5,0) is $x^{2} /4+y^{2} /21=1$.

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:

$x^{2} /a^{2} +y^{2} /b^{2} =1$,where a is the semi major axis.

We have been given that a=5, c=2.

We know that $c^{2} =a^{2} -b^{2}$

We have to find the value of b.

b=$\sqrt{a^{2} -c^{2} }$

=$\sqrt{5^{2} -2^{2} }$

=$\sqrt{25-4}$

=$\sqrt{21}$

Using the values of a and b in the equation $x^{2} /a^{2} +y^{2} /b^{2} =1$

$x^{2} /(2)^{2} +y^{2}/\sqrt{21} ^{2} =1$

$x^{2} /4+y^{2} /21=1$

Hence the equation of the ellipse having foci  (±2,0) and vertices (±5,0) is $x^{2} /4+y^{2} /21=1$.

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