old test questions i cant do. Conics (1 Viewer)
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#Edit#: On further thought, I'm a little confused at the answer I got, anyone want to correct me?
Don't take my word for it, but this is what I came up with:
Find e in terms of t using:
b^2 = a^2(e^2-1)
You should find:
e = sqrt(m+8)/sqrt(m)
Using this, you can find the foci using:
Foci = (ae,0) or (-ae,0)
You should find:
x = sqrt(2m+16)
y = 0
You will only need one equation, I'll explain why later.
Sub this point into the original equation:
3x+y-6=0
Which gives us an equation in terms of m only:
3(sqrt(2m+16))-6 = 0
Because of the Square root, you will need to square both sides after moving the 3 and the 6 to the right, therefore even if it was -ae, it would turn positive anyways.
Final Answer:
Don't take my word for it, but this is what I came up with:
Find e in terms of t using:
b^2 = a^2(e^2-1)
You should find:
e = sqrt(m+8)/sqrt(m)
Using this, you can find the foci using:
Foci = (ae,0) or (-ae,0)
You should find:
x = sqrt(2m+16)
y = 0
You will only need one equation, I'll explain why later.
Sub this point into the original equation:
3x+y-6=0
Which gives us an equation in terms of m only:
3(sqrt(2m+16))-6 = 0
Because of the Square root, you will need to square both sides after moving the 3 and the 6 to the right, therefore even if it was -ae, it would turn positive anyways.
Final Answer:
m = -6
Last edited:
alakazimmy
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b<sup>2</sup>=a<sup>2</sup>(e<sup>2</sup> - 1)
so e<sup>2</sup> = (b<sup>2</sup>/a<sup>2</sup>) + 1
= 16/m<sup>2</sup> + 1
= (16 + m<sup>2</sup>)/m<sup>2</sup>
Foci : (+ae, 0)
So, the foci are: (+ sqrt(16 + m<sup>2</sup>), 0)
Substitute this back into your line: 3x+y-6=0
Hence, 3 * sqrt (16 + m<sup>2</sup>) = 6
sqrt(16 + m<sup>2</sup>) = 2
16 + m<sup>2</sup> = 4
There is no answer for this question...?
so e<sup>2</sup> = (b<sup>2</sup>/a<sup>2</sup>) + 1
= 16/m<sup>2</sup> + 1
= (16 + m<sup>2</sup>)/m<sup>2</sup>
Foci : (+ae, 0)
So, the foci are: (+ sqrt(16 + m<sup>2</sup>), 0)
Substitute this back into your line: 3x+y-6=0
Hence, 3 * sqrt (16 + m<sup>2</sup>) = 6
sqrt(16 + m<sup>2</sup>) = 2
16 + m<sup>2</sup> = 4
There is no answer for this question...?
GUSSSSSSSSSSSSS
Active Member
wouldnt u also have to solve for cases in which a^2 = 16 and b^2 = m^2
as well as the other way round...
as well as the other way round...
a^2 is 2m, not m^2
Well not really, seeing as we have a linear equation to sub into.