write the standard form equation of an ellipse with a center at (1,2) vertex at (7,2) and focus at (4,2)

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Ответ:

Amandachavez94

11.10.2022

Mathematics

Answer option A

From the given graph is a Vertical ellipse

Center of ellipse = (-2,-3)

Vertices are (-2,3) and (-2,-9)

Co vertices are (-6,-3) and (2,-3)

The distance between center and vertices = 6, so a= 6

The distance between center and covertices = 4 , so b= 4

The general equation of vertical ellipse is

$\frac{(x-h)^2}{b^2} + \frac{(y-k)^2}{a^2}=1$

(h,k) is the center

we know center is (-2,-3)

h= -2, k = -3 , a= 6 and b = 4

The standard equation becomes

$\frac{(x+2)^2}{4^2} + \frac{(y+3)^2}{6^2}=1$

$\frac{(x+2)^2}{16} + \frac{(y+3)^2}{36}=1$

Foci are (h,k+c) and (h,k-c)

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

Plug in the a=6 and b=4

$c=\sqrt{6^2-4^2}$

$c=\sqrt{20}$

$c=2\sqrt{5}$ , we know h=-2 and k=-3

Foci are $(-2,-3+2\sqrt{5})$ and $(-2,-3-2\sqrt{5})$

Option A is correct

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