Question

Write an equation for the ellipse that satisfies the conditions: I am not sure to begin honestly; I just need a good starting point to begin.

8. endpoints of major axis (-6, 0) and (6, 0) and foci ( -sqrt 32, 0) and ( sqrt 32, 0)

Answer 1

@ParthKohli

Answer 2

`endpoints of major axis (-6, 0) and (6, 0)`

what does this tell us about the value of semi major axis length : `a` ?

Answer 3

It tells us that it goes up 6 and down -6

Answer 4

not exactly, ` (-6, 0) and (6, 0)` are points on x axis, right ?

Answer 5

left and right

Answer 6

yes that means \(\large a = 6\)

Answer 7

\(\large c = \sqrt{32}\)

Answer 8

find the value of \(\large b\) using below relation :

\[\large c^2 = a^2 - b^2\]

Answer 9

is that right b = +/- 2 @ganeshie8

Answer 10

yes, since b is a length it can only by positive so b = +2

Answer 11

can you write the equation of ellipse now ?

Answer 12

you have \(\large a = 6\) and \(\large b = 2\)
equation of ellipse would be ?

Answer 13

\[\frac{ x^2 }{ 36 } + \frac{ y^2 }{ 4 }\]

Answer 14

= 1

Answer 15

Excellent !

Answer 16

is that right?

Answer 17

Yep !
you may check with wolfram when in doubt
http://www.wolframalpha.com/input/?i=focus+x%5E2%2F36%2By%5E2%2F4%3D1

scroll down

Answer 18

what if it was switched like major axis was (0, 12) and (0, -12) and foci was at (0, sqrt 23) and (0, - sqrt 23) would the problem working be the same as the one we just did, but different answers.

Answer 19

@ganeshie8

Answer 20

yes it will be more or less same, but the most important thing is to find out whether the ellipse is horizontal or vertical

Answer 21

`major axis was (0, 12) and (0, -12) `

since the major axis is on y axis, the ellipse is VERTICAL and \(\large a = 12\)

Answer 22

\(\large c = \sqrt{23}\)


as usual find the value of \(\large b \)

Answer 23

but be careful to use the correct equation for `vertical ellipses` :
\[\large \dfrac{x^2}{b^2} + \dfrac{y^2}{a^2} = 1\]

Answer 24

\(\large a\) goes below the `y` variable ^^

Answer 25

its horizontal \[\frac{ x^2 }{ 144 } + \frac{ y^2 }{ 11 } = 1\]

Answer 26

nope, it has to be vertical because you're given :
`major axis was (0, 12) and (0, -12) `

major axis is along y axis, right ?

Answer 27

well switch 144 and 11 then

Answer 28

11 is 121

Answer 29

a = 12, b = 11

yes ! the equation would be

\[\large \frac{ x^2 }{ 121 } + \frac{ y^2 }{ 144 } = 1\]

Answer 30

so I had it right just messed up on denominator

Answer 31

yes refer to that chart, its very useful