robert7248
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What are the coordinates of the center of the ellipse shown below?

[tex]\frac{(x-7)^2}{4} + \frac{(y+3)^2}{16} = 1[/tex]

A. (-7,3)
B. (4,16)
C. (2,4)
D. (7,-3)

Respuesta :

luisejr77

Answer:

Option D (7, -3)

Step-by-step explanation:

We know that the general equation of an ellipse has the form:

[tex]\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1[/tex]

Where the point (h, k) are the coordinates of the center of the ellipse

In this case the equation of the ellipse is:

[tex]\frac{(x-7)^2}{4} + \frac{(y+3)^2}{16} = 1[/tex]

Then

[tex]h=7\\\\k = -3[/tex]

So The coordinates of the center of the ellipse are (7, -3)

lucic

Answer:

D. (7,-3)

Step-by-step explanation:

This equation is for vertical Ellipse;

For vertical Ellipse;

  • center of ellipse is given by (h,v)
  • vertices for ellipse is given by (h, v ± a)
  • co-vertices for the ellipse is given by (h ±b, v)

where the equation is (x-h)²/b² + (y-v)²/a²

In this question;

h=7 and v= -3

center= (7,-3)

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