Question

Can someone please check my work?
Write the standard form for a conic section with a center at (3,1), a vertical semi-major axis 6 units long, and e=a/3

Answer 1

\[\frac{ (X-3)^{2} }{ 9 }+\frac{ (y-1)^{2} }{ 8}=1\]

Answer 2

if e=c/a i squared a which is 3 and it is 1 then i used the formula\[b ^{2}=a ^{2}-c ^{2}\]

Answer 3

@ganeshie8

Answer 4

@ShailKumar

Answer 5

@ganeshie8 I don't know why it is liked foreign language to me. Pleeeeeeease, help

Answer 6

See, the major axis is vertical !

Answer 7

So, b > a

Answer 8

ok but when i tried to find b it is larger than a if e=c/a then a^2 must be 9 and c^6 must be 1

Answer 9

Since the ellipse is vertical, you need to change the roles of a and b.

Answer 10

Wait.. you have taken b = 3. right ?

Answer 11

oh ok well i did a=3 and a^2=9 but i guess a=3

Answer 12

I am back, hihihi.. e =a/3 so e = 6/3 =2 >1 the conic must be a hyperbola!!

Answer 13

The ellipse is \[\frac{ x ^{2} }{ a ^{2}} + \frac{ y ^{2} }{ 36} = 1\]

Answer 14

Now, e = a/3

Answer 15

so is 3=c?

Answer 16

But \[e = \sqrt{1- \frac{ a^2 }{ b^2 }}\]

Answer 17

Put b = 6 and e = a/3 and solve for a. You should get a^2 = 36/5

Answer 18

isnt it b^2 though in the formula?

Answer 19

Rephrase your doubt.

Answer 20

i am confused because 36=b^2 so b=6 right? in the formula i should then do\[\frac{ a}{ 3 }=\sqrt{1-\frac{ a ^{2} }{ 36 }}\] ?

Answer 21

how did you get 5?

Answer 22

Square both sides and solve for a^2

Answer 23

Could you get the sense of this problem ?

Answer 24

ok so a^2=7.2

Answer 25

I post my work as reference . You guys can ignore me. hihihi

Answer 26

Yes a ^2 = 7.2 but you can write the ellipse as
\[\frac{ 5(x-3)^2 }{ 36} + \frac{ (y-1)^2 }{ 36 } = 1 \ or \ 5(x-3)^2 + (y-1)^2 = 36\]

Answer 27

ok thank you so much!!!