OpenStudy (anonymous):

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.

5 years ago
OpenStudy (anonymous):

Write the equation in standard form

5 years ago
OpenStudy (anonymous):

\[ax^2+2(y^2-2y+1)-2a=0\]

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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.

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

2b^2/a

5 years ago
OpenStudy (anonymous):

values of a and b for this equation?

5 years ago
OpenStudy (anonymous):

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?

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

\[a=\sqrt{2} \ \ and \ \ b=\sqrt{a}\]

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

thats right

5 years ago
OpenStudy (anonymous):

Alright. Thank you!

5 years ago