Answer:
[tex]\frac{x^{2} }{1} +\frac{y^{2} }{2}=1[/tex]
Step-by-step explanation:
The directrices of an ellipse are defined by
[tex]x=(+-)\frac{a}{e}[/tex]
Where [tex]a >b[/tex], and [tex]e[/tex] is the eccentricity defined by [tex]e=\frac{c}{a}[/tex]
So, in the first choice, we have the equation
[tex]\frac{x^{2} }{4}+\frac{y^{2} }{1} =1[/tex]
Where [tex]a^{2}=4[/tex] and [tex]b^{2}=1[/tex], which means [tex]a=2[/tex] and [tex]b=1[/tex].
In an ellipse, we have the relation
[tex]c^{2}=a^{2}-b^{2}[/tex]
Replacing values, we have
[tex]c^{2} =4-1=3\\c=\sqrt{3}[/tex]
So, the eccentricity is [tex]e=\frac{\sqrt{3} }{2}[/tex] and the directrices are [tex]x=(+-)\frac{2}{\sqrt{3} }[/tex], which means the first choice is not the right answer. In other words, the right answer is the second choice, because third and fourth choices don't represent an ellipse, both fractions must be positive in that case.
Therefore, the right answer is
[tex]\frac{x^{2} }{1} +\frac{y^{2} }{2}=1[/tex]
Answer:
OPTION B IS CORRECT
Step-by-step explanation:
FOR A VERTICAL ELLIPSE a IS ALWAYS LARGER THAN b AND a is always right under y-squared in the equation.
