Ellipse (multiple choice)

[ZyzzBrah]

Find the equation of the ellipse with center at the origin and axis along the x-axis and one end of the latus rectum is at $\displaystyle (2 , \frac{\sqrt{5}}{5})$
Choices
a. x^2 - 4y^2 = 4
b. x^2 - 6y^2 = 6
c. x^2 + 5y^2 = 5
d. x^2 + 7y^2 = 7

 

[wjm11]

Ellipse (multiple choice)

Find the equation of the ellipse with center at the origin and axis along the x-axis and one end of the latus rectum is at

Choices
a. x^2 - 4y^2 = 4
b. x^2 - 6y^2 = 6
c. x^2 + 5y^2 = 5
d. x^2 + 7y^2 = 7

Choices "a" and "b" are not ellipses. (You can tell by the "-" sign.) Choices "c" and "d" are ellipses. You are given a point at "one end of the latus rectum." What does that mean? How can we use that info?

http://mathworld.wolfram.com/LatusRectum.html
http://hotmath.com/hotmath_help/topics/latus-rectum.html

For a good look at ellipses, visit here:

http://www.purplemath.com/modules/ellipse.htm

 

[ZyzzBrah]

how do i solve the general equation using the given?
because it leads me to a quadratic equation and having decimal numbers

e.g.
$\displaystyle LR = \frac{2\sqrt{5}}{5} = \frac{2b^2}{a}$
where $\displaystyle b^2 = a^2 - c^2$
where c = 2
substituting gives me a quadratic equation for "a" which gives a decimal number

thus i cant from the choices when i have decimal number

 

[wjm11]

how do i solve the general equation using the given?
because it leads me to a quadratic equation and having decimal numbers

e.g.

where

where c = 2
substituting gives me a quadratic equation for "a" which gives a decimal number

Was your "decimal number" equal to sqrt(5)?
If you divide the equation x^2 + 5y^2 = 5 by 5, the denominator under the x^2 is a 5. This means that a^2 = 5, or that a = sqrt(5).