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# Label the foci, the major vertices, and the minor vertices. (𝑥−1)^2/49 +(𝑦+3)^2/169 =1?

### 1 Answer

- anonymousLv 79 months agoFavorite Answer
The standard form of an equation for an ellipse with center (h, k) is

[(x - h)² / a²] + [(y - k)² / b²] = 1

Comparing that with the equation in the question,

the center is (h, k) = (1, -3)

a² = 49 , so

a = √49 = 7 …………… only interested in the positive square root here

b² = 169 , so

b = √169 = 13

b > a , so the major axis is parallel to the y-axis.

(i.e., the major axis is vertical)

For the minor axis vertices, start at the x-value of the center point (1, -3) , then go a = 7 units to both the left and right.

minor axis vertices:

( (1 - 7) , -3 ) = (-6, -3)

( (1 + 7) , -3 ) = (8, -3)

For the major axis vertices, start at the y-value of the center point (1, -3) , then go b = 13 units both upwards and downwards.

major axis vertices:

( 1 , (-3 - 13) ) = (1, -16)

( 1 , (-3 + 13) ) = (1, 10)

For an ellipse with major axis parallel to the y-axis, the foci are

(h, k - c) and (h, k + c) where c = √(b² - a²)

In this case,

c = √(13² - 7²)

= √(169 - 49)

= √120

= √(4 * 30)

= √(2² * 30)

= 2√30

so the foci are

(1, -3 - 2√30) and (1, -3 + 2√30)

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