Finding the Equation of an Ellipse given the Length of the Latus Rectum and the Distance between the Foci For a National Board Exam Review:


  Find the equation of the ellipse having a length of latus rectum of ${ \frac{3}{2} }$ and the distance between the foci is ${ 2\sqrt{13} }$


Answer is ${ \frac{x^2}{16} + \frac{y^2}{3} = 1 }$

So I try:

$${ LR = \frac{2b^2}{a} = \frac{3}{2} }$$

$${ a^2 - b^2 = c^2 }$$
$${ a^2 - b^2 = ( 2\sqrt{13} )^2 }$$

Solve two equations; I get a = 7.59...

From there I'm stuck. I cant use the variable for the answer... What is wrong with my method?
$c=13$, 
$b^2=\frac{3a}4$

Substituting in $a^2-b^2=c^2$,

$$a^2-\frac{3a}{4}=13$$

Multiply by $4$ to cancel fraction and make it whole.

$$4a^2-3a-52=0$$

$a=4$ and $a=\frac{-13}{4}$

Since we need positive use $4$ to substitute in $$\tag{1}b^2=\frac{3a}4$$

$b^2=3$ and substitute the value of $b$ in $(1)$. You can have $a^2=16$.

Therefore; 
$$\frac{x^2}{16}+\frac{y^2}{3}=1$$ or $$3x^2+16y^2=48$$
$c=13$, 
$b^2=\frac{3a}4$

Substituting in $a^2-b^2=c^2$,

$$a^2-\frac{3a}{4}=13$$

Multiply by $4$ to cancel fraction and make it whole.

$$4a^2-3a-52=0$$

$a=4$ and $a=\frac{-13}{4}$

Since we need positive use $4$ to substitute in $$\tag{1}b^2=\frac{3a}4$$

$b^2=3$ and substitute the value of $b$ in $(1)$. You can have $a^2=16$.

Therefore; 
$$\frac{x^2}{16}+\frac{y^2}{3}=1$$ or $$3x^2+16y^2=48$$