old test questions i cant do. Conics (1 Viewer)

[sincred91]

If the line 3x+y-6=0 passes through a focus of x^2/m2 - y^2/16 =1 then m equals. Please show all working.

 

[SparkyZ]

#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:

Spoiler

m = -6

 

[alakazimmy]

b²=a²(e² - 1)

so e² = (b²/a²) + 1
= 16/m² + 1
= (16 + m²)/m²

Foci : (+ae, 0)

So, the foci are: (+ sqrt(16 + m²), 0)

Substitute this back into your line: 3x+y-6=0

Hence, 3 * sqrt (16 + m²) = 6
sqrt(16 + m²) = 2
16 + m² = 4
There is no answer for this question...?

 

[GUSSSSSSSSSSSSS]

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...

 

[SparkyZ]

b²=a²(e² - 1)

so e² = (b²/a²) + 1
= 16/m² + 1
= (16 + m²)/m²

a^2 is 2m, not m^2

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...

Well not really, seeing as we have a linear equation to sub into.