Answer:
Part 1) [tex]a=\frac{b}{\sqrt{1-e^2}}[/tex]
Part 2) [tex]a=4.62\ in[/tex]
Step-by-step explanation:
The correct question in the attached figure
Part 1)
we have the formula
[tex]e=\sqrt{1-\frac{b^2}{a^2}}[/tex]
where
e the eccentricity
a is the length of the semi-major axis
b is the length of the major axis
solve for a
That means
Isolate the variable a
[tex]e=\sqrt{1-\frac{b^2}{a^2}}[/tex]
squared both sides
[tex]e^2=1-\frac{b^2}{a^2}[/tex]
subtract 1 both sides
[tex]e^2-1=-\frac{b^2}{a^2}[/tex]
Multiply both sides by a^2
[tex](e^2-1)a^2=-b^2[/tex]
Divide both sides by (e^2-1)
[tex]a^2=\frac{-b^2}{e^2-1}[/tex]
Rewrite
[tex]a^2=\frac{-b^2}{-(1-e^2)}[/tex]
[tex]a^2=\frac{b^2}{(1-e^2)}[/tex]
square root both sides
[tex]a=\sqrt{\frac{b^2}{(1-e^2)}}[/tex]
simplify
[tex]a=\frac{b}{\sqrt{1-e^2}}[/tex]
Part 2) we have
[tex]b=4\ in\\e=1/2[/tex]
Solve for a
substitute in the formula of part 1)
[tex]a=\frac{4}{\sqrt{1-(1/2)^2}}=4.62\ in[/tex]
