Find the product of the real values 'a' such that the equation $$ax^{2} + 2y^{2} - 4y + 2(1 - a) = 0$$represents an ellipse with latus rectum of length 1.

Question

anonymous · Answer

Write the equation in standard form

anonymous · Answer

$$ax^2+2(y^2-2y+1)-2a=0$$

anonymous · Answer

$$\frac{x^2}{2}+\frac{(y-1)^2}{a}=1$$

anonymous · Answer

Alright, I get how we got to standard form, but how do we find out the values of a? I got this far yesterday, but I don't know what to do from here.

anonymous · Answer

do u know what is the latus rectum formula for standard form of ellipse

anonymous · Answer

Isn't it b^2/a? I think that's what someone said when I asked it yesterday.

anonymous · Answer

2b^2/a

anonymous · Answer

values of a and b for this equation?

anonymous · Answer

Well, I don't know because isn't a the bigger one? Would it be:
4/√a = 1
a = 16
OR
2a/√2 = 1
2a = √2
a = √2/2
Then the product:
16√2/2 = 8√2
Is that right?

anonymous · Answer

if a>b latus rectum=2b^2/a
if a<b latus rectum=2a^2/b

anonymous · Answer

$$a=\sqrt{2}  \ \ and \ \ b=\sqrt{a}$$

anonymous · Answer

Ok, I figured all that stuff out before I solved for a, so did I get it right 8√2?

anonymous · Answer

thats right

anonymous · Answer

Alright. Thank you!