Question

The equation (x-7^2/64)+(y+2)^2/9=1 represents an ellipse.

Which points are the vertices of the ellipse?
(7, 1) and (7, −5)
(7, −10) and (7, 6)
(10, −2) and (4, −2)
(15, −2) and (−1, −2

Answer 1

D aka (15,-2) and (-1,-2)

Explanation:

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Answer 2

Vertices of the ellipse are, (15, -2) and (-1, -2)

Explanation:

The equation of the ellipse is,

`(\left(x-7\right)^2)/(64)+(\left(y+2\right)^2)/(9)=1`

The general equation of ellipse with centre as (h, k) is,

`((x-h)^2)/(a^2)+((y-k)^2)/(b^2)=1`

Comparing the given equation with the general form, we get the centre of the given ellipse at (7, -2) and a=8, b=3

The line through the foci intersects the ellipse at two points, the vertices.

We know that the coordinates of the vertices when (a>b) are,

`=(h\pm a,k)`

So, vertices of the given ellipse are,

`=(7+8,-2),(7-8,-2)\\\\=(15,-2),(-1,-2)`