Question

Find the equation of the ellipse coordinates of whose foci are (±2, 0) and length of latus rectum is 10 /3 .

Answer 1

then u use the formula of equation of ellicpse then u find ur equation
I hope it help u

Answer 2

Answer:

Equation can be written as

$\frac{x^2}{\frac{35+\sqrt{41}}{2}}+\frac{y^2}{\frac{25+5\sqrt{41}}{2}}=1$

Step-by-step explanation:

Given the coordinates of foci  (±2, 0) and length of latus rectum is $\frac{10}{3}$

We have to find the equation of ellipse.

As the foci lies on x-axis

Equation will be

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Here, c=ae=2

⇒ $e=\frac{2}{a}$

$\text{Length of latus rectum=}\frac{2b^2}{a}=10$

$b^2=5a$

⇒ $a^2(1-e^2)=5a$

⇒  $a(1-e^2)=5$

⇒   $a(1-\frac{4}{a^2})=5$

⇒ $a^2-4=5a$

⇒ $a^2-5a-4=0$

$a=\frac{5\pm\sqrt{5^2-4(1)(-4)}}{2(1)}=\frac{5\pm\sqrt{41}}{2}$

But a>0

⇒ $a^2=(\frac{5+\sqrt{41}}{2})^2=\frac{25+41+10\sqrt{41}}{4}=\frac{35+5\sqrt{41}}{2}$

$b^2=5(\frac{5+\sqrt{41}}{2})=\frac{25+5\sqrt{41}}{2}$

Hence, equation can be written as

$\frac{x^2}{\frac{35+5\sqrt{41}}{2}}+\frac{y^2}{\frac{25+5\sqrt{41}}{2}}=1$