(A) The vertices are $(\pm 5, 0)$,which lie on the $x$-axis. Thus,the equation is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Here,$a = 5$ and

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$\frac{x^2}{25} + \frac{y^2}{9} = 1$

(A) The vertices are $(\pm 5, 0)$,which lie on the $x$-axis. Thus,the equation is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Here,$a = 5$ and the distance of the foci from the center is $c = 4$.Using the relation $a^2 = b^2 + c^2$,we have $5^2 = b^2 + 4^2$.$25 = b^2 + 16 \implies b^2 = 9$.Substituting $a^2 = 25$ and $b^2 = 9$ into the standard equation,we get $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

Class 11 Mathematics 10-2. Parabola, Ellipse, Hyperbola Ellipse

Medium

Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 5, 0)$,foci $(\pm 4, 0)$.

A
$\frac{x^2}{25} + \frac{y^2}{9} = 1$
B
$\frac{x^2}{9} + \frac{y^2}{25} = 1$
C
$\frac{x^2}{16} + \frac{y^2}{25} = 1$
D
$\frac{x^2}{25} + \frac{y^2}{16} = 1$

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10-2. Parabola, Ellipse, Hyperbola

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Find the equation for the ellipse that satisfies the given conditions: Vertices $(\pm 5, 0)$,foci $(\pm 4, 0)$.

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